Answer:
HgBr2 conducts when molten because there are mobile ions in molten HgBr which allows flow of current when an electrical potential difference is introduced to the HgBr in molten state
However HgBr2 does not conduct in the solid state as the ions are fixed in the solid HgBr2 lattice structure
Mercury, which is a metal in its natural form conducts both in the solid and molten states as the delocalized electrons are able to move both in the solid and molten mercury states and as such current flows through mercury when there is an electrical potential difference placed across it
Explanation:
Electricity or electric current flow is the term used to describe the state of movement or flow of matter that carries an electrical charge
It is the steady movement of or flow of electrons. The moving electrons transfer electrical charge round an electrical circuit. In metals, there are freely shared electrons between individual atoms so as to efficiently conduct electricity and so when an electrical potential difference is placed across a piece of metallic object an electron is readily displaced by another electron entering from one end and exiting from the other end of the electrical potential difference
Mercury bromide and mercury exhibit electrical conductivity in both solid and molten states due to the presence of freely mobile, charged entities. In the solid state, ionic compounds like mercury bromide are not conductive, but they become conductive when molten.
Explanation:In both the solid and molten states, mercury bromide and mercury demonstrate electrical conductivity due to the presence of freely mobile, charged entities. In the solid state, ionic compounds like mercury bromide are not electrically conductive because their ions are unable to flow. However, in the molten state, the ions are able to move freely through the liquid, allowing for electrical conductivity.
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An element has four naturally occurring isotopes with the masses and natural abundances given in the table below. Find the atomic mass of the element (express your answer to four significant figures and include the appropriate units).
Isotope Mass (amu) Abundance (%)
1 203.97304 1.390
2 205.97447 24.11
3 206.97590 22.09
4 207.97665 52.41
Identify the element, by spelling out its full name.
Answer:
Lead (Pb)
Explanation:
To find the atomic mass of the element, we need to take into consideration all the naturally occurring isotopes. We then find the atomic mass by multiplying the natural abundance of the isotopes by their mass for all of the isotopes and summing them together.
1. 1.390/100 * 203.97304 = 2.835225256
2. 24.11/100 * 205.97447 = 49.660444717
3. 22.09/100 * 206.97590 = 45.72097631
4. 52.41/100 * 207.97665 = 109.000562265
We then add all of these masses together:
109.000562265 + 45.72097631+ 49.660444717 + 2.835225256 = 207.217208548
To 4 sf = 207.2 amu
Element is Lead (Pb)
Which of the following would be expected to have the lowest freezing point? a. 0.1 M NaCl b. 0.1 M MgCl2 c. 0.1 M AlCl3
Answer:
option c, 0.1 M [tex]AlCl_3[/tex]
Explanation:
Addition of non-volatile solute to a solvent decreases its vapour pressure which results in decrease in melting point.
Decrease in melting point is known as depression in freezing point. the depression in freezing point is related with molality and no. of ions as follows:
[tex]\Delta T_f = imK_f[/tex]
Where, i is von't Hoff factor, m is molatilty and ΔTf is depression in freezing point.
As, in the given case concentration of all the solution is same, therefore, depression in freezing point will depend upon the no. of ions produced by the ionization of the salts in the aqueous solution.
In case of NaCl, the no. of ions produced will be 2.
Therefore, value of i will be 2
In case of [tex]MgCl_2[/tex], the no. of ions produced will be 3.
Therefore, value of i will be 3
In case of [tex]AlCl_3[/tex], the no. of ions produced will be 4.
Therefore, value of i will be 4.
More, the value of van't Hoff factor, more will be depression in freezing point.
Therefore, assuming the given solution to be aqueous, the solution expected to have lowest freezing point is 0.1 M [tex]AlCl_3[/tex].
The correct option is b. 0.1 M MgCl2 would be expected to have the lowest freezing point.
To understand why 0.1 M MgCl2 has the lowest freezing point, we need to consider the colligative properties of solutions, specifically the freezing point depression. The freezing point depression is directly proportional to the number of particles (ions or molecules) in the solution. When a solute is dissolved in a solvent, it dissociates into ions, and the more ions it dissociates into, the greater the freezing point depression.
Let's analyze each of the given solutions:
a. 0.1 M NaCl: Sodium chloride (NaCl) dissociates into two ions in solution, Na+ and Cl-. Therefore, the total number of particles in solution is 2 times the concentration of NaCl, which is 0.2 moles of particles per liter.
b. 0.1 M MgCl2: Magnesium chloride (MgCl2) dissociates into three ions in solution, Mg2+ and 2Cl-. Therefore, the total number of particles in solution is 3 times the concentration of MgCl2, which is 0.3 moles of particles per liter.
c. 0.1 M AlCl3: Aluminum chloride (AlCl3) can dissociate into four ions in solution, Al3+ and 3Cl-. However, AlCl3 is not fully dissociated in aqueous solution due to its Lewis acid behavior and the formation of complex ions such as AlCl4-. Assuming complete dissociation for simplicity, which is not the case in reality, the total number of particles in solution would be 4 times the concentration of AlCl3, which is 0.4 moles of particles per liter.
Since MgCl2 dissociates into the most number of ions compared to NaCl and AlCl3 (assuming complete dissociation for AlCl3), it will have the greatest effect on freezing point depression. Therefore, 0.1 M MgCl2 is expected to have the lowest freezing point among the given options.
It is important to note that in reality, the actual freezing point depression of AlCl3 would be less than calculated here due to its incomplete dissociation and complex ion formation. This further supports the conclusion that MgCl2 would have the lowest freezing point, as AlCl3 would not reach the expected freezing point depression based on complete dissociation.
When aluminum hydroxide (dissolved in water);is mixed with aqueous sulfuric acid (H2SO4) the products are aluminum sulfate (a precipitate) and liquid water
Answer:
The equation for this reaction is:
2Al(OH)3 (aq) + 3H2SO4 (l) ---> Al2(SO4)3 (s) + 6H2O (l)
Explanation:
At − 12.0 ∘ C , a common temperature for household freezers, what is the maximum mass of sorbitol (C6H14O6) you can add to 2.00 kg of pure water and still have the solution freeze? Assume that sorbitol is a molecular solid and does not ionize when it dissolves in water.
Answer:
2,347.8 grams
Explanation:
The freezing point depression Kf of water = 1.86° C / molal.
To still freeze at -12° C, then the molality of the solution 12/ 1.86 = 6.45 moles
The molecular weight of sorbitol (C6H14O6)is:
6 C = 6 ×12 = 72
14 H = 14 × 1 = 14
6 O = 6 × 16 = 96
...giving a total of 182
So one mole of sorbitol has a mass of 182 grams.
Since there are 2 kg of water, 2 × 6.45 moles = 12.9 moles can be added to the water to get the 12° C freezing point depression.
Therefore
grams = moles × molar mass
12.9 moles × 182 grams / mole = 2,347.8 grams of sorbitol can be added and still freeze
What happens when you add more solute to a saturated solution
Answer:
If you add more solute in a saturated solution, it will have no effect. Additional solute does not dissolve in a saturated solution.
Explanation:
If more solute is added and it does not dissolve, then the original solution was saturated. If the added solute dissolves, then the original solution was unsaturated. A solution that has been allowed to reach equilibrium but which has extra undissolved solute at the bottom of the container must be saturated.
Given a unsorted list of 1024 elements, what is the runtime for linear search if the search key is less than all elements in the list?
Answer:
10
Explanation:
Binary search's runtime is proportional to log (base two) of the number of list elements.
In the construction industry, why are I beams generally used as support as opposed to solid rectangular beams? What might be an advantage of using a rectangular beam?
Answer:I beams can withstand greater Moment compared to other beams like rectangular beam when given the same stress.
Rectangular beams are better for casting concrete beams, and for making beams of high and low moment of inertia and its centriod is easy to understand.
Explanation: I beams are beams usually used in metal work and in making beams of high moments of inertia.
Rectangular beams are beams whose centroid ( the geometric center of a regular rectangular beam) are easily identifiable and they can be built to give high and low moment of inertia. This is one of the advantage of using rectangular beams.
The advantage of use of I beam over rectangular beam has been the wide spread and high moment of inertia.
In the construction industry, for the support to the buildings and the structure, beams of different shapes has been used. I beam are the steel or metal beams that have high functionality.
I beam in construction industryI beam has been made with the shape of rolled joist. They have been installed in the building because of their high moment of inertia.
I beam has been working with the spread of moment of inertia in the structure, that has been able to bend and withstand the pressure more than compared to the rectangular cross-section with concentrated inertia.
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A uniform, solid, 1100.0 kg sphere has a radius of 5.00 m. Find the gravitational force this sphere exerts on a 2.40 kgkg point mass placed at the following distances from the center of the sphere: (a) 5.01 mm , and (b) 2.50 mm .
Explanation:
(a) The given data is as follows.
M = 1100 kg, r = 5.01 mm = [tex]5.01 \times 10^{-3}[/tex] m
m = 2.40 kg
Now, formula for Newton's law of gravitation is as follows.
F = [tex]\frac{GMm}{r^{2}}[/tex]
Putting the given values into the above formula as follows.
F = [tex]\frac{GMm}{r^{2}}[/tex]
= [tex]\frac{6.673 \times 10^{-11} Nm^{2}/kg^{2} \times 1100 \times 2.40 kg}{(5.01 \times 10^{-3})^{2}}[/tex]
= [tex]7.02 \times 10^{-3}[/tex] N
Therefore, gravitational force exerted by the given sphere on 2.40 kg point mass is [tex]7.02 \times 10^{-3}[/tex] N.
(b) Now, mass of the sphere beyond r = 2.50 mm or [tex]2.50 \times 10^{-3} m[/tex] will not contribute any force on the point mass.
Hence, mass within the radius r < R that experiences force is,
[tex]m^{1} = M \frac{r^{3}}{R^{3}}[/tex]
According to Newton's law of gravitation,
F = [tex]\frac{Gm^{1}m}{r^{2}}[/tex]
= [tex]\frac{Gm(M \frac{r^{3}}{R^{3}})}{r^{2}}[/tex]
= [tex]G \frac{mMr}{R^{3}}[/tex]
Here, r is the radius of point mass and R is the radius of solid sphere.
Therefore, putting the given values into the above formula as follows.
F = [tex]G \frac{mMr}{R^{3}}[/tex]
= [tex]\frac{6.673 \times 10^{-11} Nm^{2}/kg^{2} \times 1100 kg \times 2.40 kg \times 2.50 \times 10^{-3}}{(5)^{3}}[/tex]
= [tex]3.52 \times 10^{-9}[/tex] N
Therefore, the gravitational force this sphere exerts on a 2.40 kg is [tex]3.52 \times 10^{-9}[/tex] N.
Analysis of a volatile liquid showed that it is 54.5% carbon, 9.1% hydrogen, and 36.4% oxygen by mass. A separate 0.345-gram sample of its vapor occupied 120. mL at 100.°C and 1.00 atm. What is the molecular formula for the compound?
Answer: The molecular formula for the given organic compound is [tex]C_4H_8O_2[/tex]
Explanation:
We are given:
Percentage of C = 54.5 %
Percentage of H = 9.1 %
Percentage of O = 36.4 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 54.5 g
Mass of H = 9.1 g
Mass of O = 36.4 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{54.5g}{12g/mole}=4.54moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{9.1g}{1g/mole}=9.1moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{36.4g}{16g/mole}=2.28moles[/tex]
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.28 moles.
For Carbon = [tex]\frac{4.54}{2.28}=1.99\approx 2[/tex]
For Hydrogen = [tex]\frac{9.1}{2.28}=3.99\approx 4[/tex]
For Oxygen = [tex]\frac{2.28}{2.28}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : O = 2 : 4 : 1
Hence, the empirical formula for the given compound is [tex]C_2H_{4}O_1=C_2H_4O[/tex]
Mass of empirical formula = [tex]C_2H_4O[/tex] = 2(12) + 4(1) + 16 = 44 g/eq.
Now we have to determine the molar mass of compound by using ideal gas equation.
[tex]PV=nRT\\\\PV=\frac{w}{M}RT[/tex]
where,
P = pressure of gas = 1.00 atm
V = volume of gas = 120 mL = 0.120 L
T = temperature of gas = [tex]100^oC=273+100=373K[/tex]
w = mass of gas = 0.345 g
M = molar mass of gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the above formula, we get:
[tex]PV=\frac{w}{M}RT[/tex]
[tex](1.00atm)\times (0.120L)=\frac{0.345g}{M}\times (0.0821L.atm/mol.K)\times (373K)[/tex]
[tex]M=88.04g/mol[/tex]
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is :
[tex]n=\frac{\text{molecular mass}}{\text{empirical mass}}[/tex]
We are given:
Mass of molecular formula = 88.04 g/mol
Mass of empirical formula = 44 g/mol
Putting values in above equation, we get:
[tex]n=\frac{88.04g/mol}{44g/mol}=2[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_2H_4O=(C_2H_4O)_n=(C_2H_4O)_2=C_4H_8O_2[/tex]
Thus, the molecular formula for the given compound is [tex]C_4H_8O_2[/tex]
The empirical formula of the given compound is [tex]\bold {C_2H_4O}[/tex]. The empirical formula is the smallest whole-number ratio of the compound.
Assume the mass of the compound is 100 g. So, the percentages given are taken as mass.
Mass of C = 54.5 g = 4.54 moles
Mass of H = 9.1 g = 9.1 moles
Mass of O = 36.4 g = 2.28 moles
To formulate the empirical formula, Calculate the molar ratio by dividing moles by the smallest number,
For the mole ratio, we divide each value of the moles by the smallest number of moles, we get.
For Carbon = 2
For Hydrogen =4
For Oxygen = 1
Therefore, the empirical formula of the given compound is [tex]\bold {C_2H_4O}[/tex].
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The observation that 4.0 g of hydrogen reacts with 32.0 g of oxygen to form a product with O:H mass ratio-8:1, and 6.0 g of hydrogen reacts with 48.0 g of oxygen to form the same product with O/H mass ratio = 8:1 is evidence for the law of 1. multiple proportions 2. erergy conservation. 3. mass conservation 4. definite proportion
Answer:
Law of definite proportion
Explanation:
As per law of definite proportion, ratio of elements present in a compound is always fixed irrespective of the source, amount and method of preparation.
In the given case, the hydrogen and oxygen react with each other to form a compound. The ratio of oxygen and hydrogen is fixed which is 8 : 1 and this ratio does not change upon changing amount of oxygen and hydrogen.
So, the this experiment support the law of definite proportion.
Therefore, the correct option is option 4
Suppose a 1.30 g nugget of pure gold has zero net charge. What would be its net charge after it has 1.68% of its electrons removed?
Answer:
The net charge of 1.3 g nugget of pure gold after 1.68% of its electrons are removed is 559 C
Explanation:
When an atom gains electrons it becomes negatively charged. Conversely, when it looses electrons the atoms becomes positively charged thus
To solve this question, we rely on the relationship between the nmber of particles present in a given mass of an atom, Avogadro's number and number of moles, n
The given variables are
mass of pure gold nugget = 1.30 g
Quantity of electrons removed = 1.68% of electrons present in the gold sample
Molar mass of gold = 197 g/mol
Avogadro's number = 6.02 × 10²³ atoms/mole
qc = one electron charge = -1.06 × 10⁻¹⁹ C/electron
Electrical charge of gold nugget = 0 C
Number of electrons in one gold atom = 79 electrons
Solving for the number of prticles or gold atoms in 1.3 grams of gold we get
n mass/(molar mass) = 1.3/197 moles of gold = 0.0066 moles
number of particles in 0.0066 moles of gold N = n×[tex]N_{A}[/tex] = 0.0066 × 6.02 × 10²³ = 3.97 × 10²¹ atoms
since 79 electrons are present per particle we have
3.97 × 10²¹ × 79 = 3.14 × 10²³ electrons
quantity of elecrtrons removed = 1.68% of 3.14 × 10²³ electrons =1.68/100 × 3.14 × 10²³ electrons = 0.0168 × 3.14 × 10²³ electrons = 5.3 × 10²¹ electrons
The net charge of 5.3 × 10²¹ electrons = 5.3 × 10²¹ electrons × -1.06 × 10⁻¹⁹ C/electron =
5.59 × 10² C = 559 C
Final answer:
After 1.68% of its electrons are removed, a 1.30 g nugget of pure gold will have a net positive charge of approximately +842 C, calculated based on the initial number of electrons and the charge per electron.
Explanation:
To calculate the net charge of a 1.30 g nugget of pure gold after 1.68% of its electrons are removed, we first need to understand the composition and characteristics of gold. The atomic mass of gold (Au) is approximately 197 g/mol, and each atom has 79 protons and, when neutral, 79 electrons. The number of gold atoms in the nugget can be calculated using its mass and the atomic mass of gold.
First, calculate the number of moles of gold in the nugget:
Moles of gold =
(1.30 g) / (197 g/mol) = 0.0066 mol. The number of gold atoms = 0.0066 mol ×
(6.022 × 1023 atoms/mol) ≈ 3.97 × 1021 atoms. Since each gold atom has 79 electrons when neutral, the total number of electrons initially is approximately 3.13 × 1023 electrons.
To find out how many electrons 1.68% represents:
1.68% of the total electrons = 1.68 / 100 × 3.13 × 1023 ≈ 5.26 × 1021 electrons.
The charge of one electron is approximately -1.6 × 10-19 C. Therefore, removing 5.26 × 1021 electrons means the nugget would have a net positive charge equal to:
5.26 × 1021 × 1.6 × 10-19 C ≈ 8.42 × 102 C. Thus, the net charge of the gold nugget after removing 1.68% of its electrons would be approximately +842 C.
When 0.5000 grams of an unknown hydrocarbon, CxHy, is completely combusted with excess oxygen, 1.037 L CO2 gas and is produced at 98.3 °C and 1.000 atm. What is the empirical formula of the hydrocarbon? (R = 0.08206 L×atm/mol×K)
Answer: C₃H₈
Explanation:
From PV = nRT
Moles of CO2 = PV / RT =(1 x 1.037)/(0.08206 x ( 273+98.3)) = 0.0340 moles of CO2
CxHy + (x + (y/2))O₂ ----> xCO₂ + yH₂O
there is 1 mole of C in each mole of CO2 so moles of C in CO₂ = 0.0340 moles
mass of C in CO₂ = 0.0340 x 12 = 0.41 g
This means mass of C in the hydrocarbon = 0.41g
so mass of H in the hydrocarbon = 0.50 - 0.41= 0.09 g
moles of H in the hydrocarbon = mass/molar mass = 0.09/1 = 0.09 moles
molar ratio of C:H = 0.034 : 0.09 = 1 :2.67
or 3 :8
empirical formula is C3H8
To find the empirical formula of the hydrocarbon, calculate the moles of carbon dioxide produced using the ideal gas law and the moles of carbon and hydrogen in the hydrocarbon. The empirical formula is C12H1.
Explanation:In order to determine the empirical formula of the hydrocarbon, we need to find the mole ratio of carbon to hydrogen in the compound. To do this, we first calculate the moles of carbon dioxide (CO2) produced using the ideal gas law:
n = (PV)/(RT) = (1.000 atm)(1.037 L)/(0.08206 L.atm/mol.K)(98.3 + 273.15 K) = 0.0443 mol
Since the combustion of 1 mole of hydrocarbon produces 1 mole of CO2, we can conclude that the moles of hydrocarbon consumed is also 0.0443 mol. Next, we calculate the moles of carbon and hydrogen in the hydrocarbon:
moles of carbon = (mass of carbon in hydrocarbon) / (molar mass of carbon) = (12.01 g/mol)(0.5000 g) / (1 g) = 6.003 mol
moles of hydrogen = (mass of hydrogen in hydrocarbon) / (molar mass of hydrogen) = (1.008 g/mol)(0.5000 g) / (1 g) = 0.5008 mol
The empirical formula of the hydrocarbon is then C6H0.5008. We can simplify this ratio by multiplying all the subscripts by 2 (the smallest whole number ratio) to get the empirical formula: C12H1.
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In nature, the element X consists of two naturally occurring isotopes. 107X with abundance 53.84% and isotopic mass 106.9051 amu and 109X with isotopic mass 108.9048 amu. Use the given information to calculate the atomic mass of the element X to an accuracy of .001% (Report your answer like this yyy.yyyy)
The atomic mass of element X is approximately 107.8682 amu with an accuracy of .001%.
To calculate the atomic mass of element X, we need to use the given information about the isotopes of element X.
We are given that element X consists of two naturally occurring isotopes: 107X with an abundance of 53.84% and an isotopic mass of 106.9051 amu, and 109X with an isotopic mass of 108.9048 amu.
To calculate the atomic mass of element X, we can use the formula:
Atomic mass = (abundance of isotope 1 x mass of isotope 1) + (abundance of isotope 2 x mass of isotope 2)
Plugging in the values for element X, we get:
Atomic mass = (0.5384 x 106.9051 amu) + (0.4616 x 108.9048 amu)
Atomic mass ≈ 107.8682 amu
Therefore, the atomic mass of element X is approximately 107.8682 amu to an accuracy of .001%.
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To calculate the atomic mass of element X, multiply the mass of each isotope by its abundance and sum them up. The atomic mass of element X is 106.22288 amu.
Explanation:To calculate the atomic mass of element X, we need to consider the abundance and isotopic mass of its two naturally occurring isotopes. The atomic mass is calculated by multiplying the mass of each isotope by its abundance and summing them up.
For isotope 107X with an abundance of 53.84%, we multiply its mass (106.9051 amu) by its abundance (0.5384). For isotope 109X with an abundance of (100% - 53.84% = 46.16%), we multiply its mass (108.9048 amu) by its abundance (0.4616).
Finally, we add these two values together to get the atomic mass of element X to an accuracy of .001%. The calculation is as follows:
(106.9051 amu * 0.5384) + (108.9048 amu * 0.4616) = 106.22288 amu
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The element gallium has an atomic weight of 69.7 and consists of two stable isotopes gallium -69 and gallium -71. The isotope gallium -69 has a mass of 68.9 amu and a percent natural abundance of 60.4 %. The isotope gallium -71 has a percent natural abundance of 39.6%. What is the mass of gallium -71?
Answer:
The answer to your question is Gallium-71 = 70.9202 amu
Explanation:
Gallium atomic weight = 69.7
Gallium-69 = 68.9 amu abundance = 60.4%
Gallium-71 = x abundance = 39.6%
To solve this problem just write an equation and solve it for the mass of gallium-71.
Equation
Gallium = Gallium-69(abundance) + Gallium-71(abundance)
Substitution
69.7 = (68.9)(0.604) + Gallium-71(0.396)
69.7 = 41.6156 + Gallium-71(0.396)
Gallium-71(0.396) = 69.7 - 41.6156
Gallium-71(0.396) = 28.0844
Gallium-71 = 28.0844/0.396
Gallium-71 = 70.9202 amu
if you begin with 210.3 g of Cl2, how many grams of HCl will you end up with? CH4 + 3Cl2 → CHCl3 + 3HCl *
Answer:
108.04g
Explanation:
Looking at the balanced chemical equation theoretically, we can see that 3 moles of chlorine yielded 3 moles of HCl. This means that they have equal mole ratio.
Now let’s get the actual reactive moles. The number of moles is obtained by dividing the mass by the molar mass.
The molecular mass of chlorine gas is (2 * 35.5) = 72g/mol
The number of moles is thus 210.3/71 = 2.96 moles
Since the number of moles are equal, the number of moles of HCl produced too is 2.96 moles.
Now to get the mass of HCl produced, we multiply the number of moles by the molar mass of HCl. The molar mass of HCl is 36.5g/mol
The mass is thus 2.96 * 36.5 = 108.04g
Given a list of 10,000 elements, and if each comparison takes 2 µs, what is the fastest possible runtime for linear search?
Answer:
2 µs
Explanation:
Nitrogen dioxide (NO2) cannot be obtained in a pure form in the gas phase because it exists as a mixture of NO2 and N2O4. At 26°C and 0.80 atm, the density of this gas mixture is 2.2 g·L−1. What is the partial pressure of each gas?
Answer: The partial pressure of [tex]NO_2[/tex] is 0.426 atm and that of [tex]N_2O_4[/tex] is 0.374 atm
Explanation:
Assuming ideal gas behavior, the equation follows:
PV = nRT
We know that:
[tex]\text{Density}=\frac{\text{Mass}}{\text{Volume}}[/tex]
Rearranging the above equation:
[tex]M=\frac{dRT}{P}[/tex]
where,
d = density of gas mixture = 2.2 g/L
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature = [tex]26^oC=[26+273]K=299K[/tex]
P = pressure of the mixture = 0.80 atm
M = average molar mass of mixture
Putting values in above equation, we get:
[tex]M_{avg}=\frac{2.2g/L\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 299}{0.80atm}\\\\M_{avg}=67.5g/mol[/tex]
We know that:
Molar mass of [tex]NO_2[/tex] = 46 g/mol
Molar mass of [tex]N_2O_4[/tex] = 92 g/mol
Let the mole fraction of [tex]NO_2[/tex] be 'x' and that of [tex]N_2O_4[/tex] be '(1-x)'
For average molar mass calculation:
[tex]M_{avg}=M_{NO_2}\chi_{NO_2}+M_{N_2O_4}\chi_{N_2O_4}[/tex]
Putting values in above equation:
[tex]67.5=46x+92(1-x)\\\\x=0.533[/tex]
Mole fraction of [tex]N_2O_4[/tex] = (1 - x) = (1 - 0.533) = 0.467
To calculate the partial pressure, we use the equation given by Raoult's law, which is:
[tex]p_{A}=p_T\times \chi_{A}[/tex]
For [tex]NO_2[/tex] :We are given:
[tex]p_T=0.80atm\\\chi_{NO_2}=0.533[/tex]
Putting values in above equation, we get:
[tex]p_{NO_2}=0.80atm\times 0.533\\\\p_{NO_2}=0.426atm[/tex]
For [tex]N_2O_4[/tex] :We are given:
[tex]p_T=0.80atm\\\chi_{N_2O_4}=0.467[/tex]
Putting values in above equation, we get:
[tex]p_{N_2O_4}=0.80atm\times 0.467\\\\p_{N_2O_4}=0.374atm[/tex]
Hence, the partial pressure of [tex]NO_2[/tex] is 0.426 atm and that of [tex]N_2O_4[/tex] is 0.374 atm
Boston, MA and Barcelona, Spain have very similar latitudes (they are both a similar distance from the equator.) Why is the climate of Boston so different from the climate of Barcelona?
Answer:
The climate of Boston is so different from the climate of Barcelona because regional climate does not depend only on latitude. The flow of ocean current and its temperature causes this different climates. East coast cities such as Boston have a continental climate because prevailing westerly winds continually bring them under the influence of continental air masses while westerlies bring maritime air masses and therefore a maritime climate to the coast cities such as Barcelona and others.
A. After the label fell off a bottle containing a clear liquid believed to be benzene, a chemist measured the density of the liquid to verify its identity. A 25.0-mL portion of the liquid had a mass of 21.95 gg. A chemistry handbook lists the density of benzene at 15∘C∘C as 0.878 g/mg/mL. Is the calculated density in agreement with the tabulated value?
B. An experiment requires 15.0 g of cyclohexane, whose density at 25oC is 0.7781 g/mL. What volume of cyclohexane should be used?
C. A spherical ball of lead has a diameter of 5.0 cm. What is the mass of the sphere if lead has a density of 11.34 g/cm3?
Answer:
A. Yes, the calculated density in agreement with the tabulated value.
B. 19.28 mL of volume of cyclohexane should be used.
C. 742.20 is the mass of the sphere of lead.
Explanation:
A.
Volume of the liquid = V = 25.0 mL
Mass of the liquid = m = 21.95 g
Density of the liquid = d
[tex]d=\frac{m}{V}[/tex]
[tex]=\frac{21.95 g}{25.0 mL}=0.878 g/mL[/tex]
Density mentioned in the report book = d' = 0.878 g/mL
d' = d = 0.878 g/mL
Yes, the calculated density in agreement with the tabulated value.
B.
Volume of the liquid cyclohexane= V = ?
Mass of the liquid cyclohexane= m = 15.0 g
Density of the liquid cyclohexane = d = 0.7781 g/mL
[tex]d=\frac{m}{V}[/tex]
[tex]V=\frac{15.0 g}{0.7781 g/mL}=19.28 mL[/tex]
19.28 mL of volume of cyclohexane should be used.
C.
Diameter of the ball = d = 5.0 cm
Radius of the ball = r = 0.5 × d = 2.5 cm
Volume of sphere ,V= [tex]\frac{4}{3}\pi r^3[/tex]
[tex]V = \frac{4}{3}\times 3.14\times (0.25 cm)^3=65.45 cm^3[/tex]
Volume of the spherical lead ball = V
Mass of the spherical lead ball= m = ?
Density of the spherical lead ball = d = [tex]11.34 g/cm^3[/tex]
[tex]d=\frac{m}{V}[/tex]
[tex]m=d\times V=11.34 g/cm^3\times 65.45 cm^3=742.20 g[/tex]
742.20 is the mass of the sphere of lead.
A. The tabulated density of benzene [tex]15^0 C[/tex] is 0.878 g/mL. B. 19.28 mL of volume of cyclohexane should be used. C. The mass of the lead sphere is approximately 743.5 g.
A. To determine if the calculated density of the liquid matches the tabulated value for benzene, we need to calculate the density using the provided data and then compare it to the handbook value. The density is calculated by dividing the mass of the liquid by its volume.
Given:
Mass of the liquid = 21.95 g
The volume of the liquid = 25.0 mL
Calculated density = [tex]Mass / Volume = 21.95 g / 25.0 mL = 0.878 g/mL[/tex]
The tabulated density of benzene [tex]15^0 C[/tex] is 0.878 g/mL. Since the calculated density matches the tabulated value, the liquid in the bottle is likely benzene.
B. To find the volume of cyclohexane required for the experiment, we can use the formula: Volume = Mass / Density
Given:
Mass of cyclohexane needed = 15.0 g
Density of cyclohexane at [tex]25^0C[/tex] = 0.7781 g/mL
Volume of cyclohexane = 15.0 g / 0.7781 g/mL = 19.28 mL
Therefore, 19.28 mL of cyclohexane should be used for the experiment.
C. The mass of the lead sphere can be calculated using the formula for the volume of a sphere and the density of lead:
The volume of a sphere = [tex](4/3)\pi r^3[/tex]
Density = Mass / Volume
Given:
The diameter of the lead sphere = 5.0 cm
Radius of the lead sphere = Diameter / 2 = 5.0 cm / 2 = 2.5 cm
Density of lead = [tex]11.34 g/cm^3[/tex]
The volume of the lead sphere =[tex](4/3)\pi (2.5 cm)^3 =20.94375\pi cm^3[/tex]
Mass of the lead sphere = Density — Volume = [tex]11.34 g/cm^3 - 20.94375\pi cm^3= 11.34 g/cm^3- 65.45902 cm^3= 743.5 g[/tex]
Therefore, the mass of the lead sphere is approximately 743.5 g.
If the rate of evaporation is equal to the rate of condensation, the system is in a state of dynamic equilibrium, which cannot be disturbed. TRUE FALSE
Answer: False
Explanation:
From definition dynamic equilibrium is a state of balance between continuing processes. When the rate of evaporation equals the rate of condensation the system has reached a dynamic equilibrium.
However, it is possible, to disturb a system that is in dynamics equilibrium by changing conditions of the system. In fact anything that changes the thermodynamic state of the system will disturb the system and it will no longer be at equilibrium.
An increase in temperature to the system will favour evaporation more than condensation. Molecules with higher kinetic energy will escape the system.
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Which of the following is true of greenhouse gases?
They exist in fixed quantities.
They reflect incoming solar radiation.
They trap energy in the atmosphere.
They are all naturally occurring.
Greenhouse gases trap energy in the atmosphere.
Explanation:
Greenhouses gases trap the radiation, especially long infrared wavelengths, that emanate from the earth's surface after solar radiation hits the earth’s surface. These infrared wavelengths are electromagnetic radiation that are significant in heat transfer. Mostly a higher percentage of this radiation from the earth’s surface escapes into space. Trapping this radiation in the atmosphere causes warming of the atmosphere - a phenomenon called greenhouse effect. This is why an increase in greenhouse gases causes an increase in global temperatures and subsequently drives climate change.
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As a scuba diver descends under water, the pressure increases. At a total air pressure of 2.71 atm and a temperature of 25.0 C, what is the solubility of N2 in a diver's blood?
[Use the value of the Henry's law constant k calculated , 6.26 x 10^{-4} (mol/(L*atm).]
Assume that the composition of the air in the tank is the same as on land and that all of the dissolved nitrogen remains in the blood.
Express your answer with the appropriate units.
Answer:
The molar solubility of nitrogen gas is [tex]1.32\times 10^{-3}mol/L[/tex].
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{N_2}=K_H\times p_{liquid}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]6.26\times 10^{-4}mol/L.atm[/tex]
[tex]p_{N_2}[/tex] = partial pressure of nitrogen gas
Total air pressure = P = 2.71 atm
Percentage of nitrogen in air = 78.09%
Mole fraction of nitrogen ,[tex]\hi_{N_2}= 0.7809[/tex]
[tex]p_{N_2}=P\times \chi_{N_2}=2.71 atm\times 0.7809[/tex]
Putting values in above equation, we get:
[tex]C_{N_2}=6.26\times 10^{-4}mol/L.atm\times 2.71 atm\times 0.7809\\\\C_{N_2}=1.32\times 10^{-3}mol/L[/tex]
Hence, the molar solubility of nitrogen gas is [tex]1.32\times 10^{-3}mol/L[/tex].
The solubility of N₂ in a diver's blood at 2.71 atm pressure and 25.0 C temperature is calculated using Henry's law to be 1.323 x 10⁻³ mol/L, assuming a mole fraction of N₂ of 0.78.
The solubility of N₂ in a diver's blood at a total air pressure of 2.71 atm and a temperature of 25.0 C can be determined using Henry's law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid. In this case, the Henry's law constant k for N₂ is given as 6.26 x 10⁻⁴ mol/(L·atm).
Assuming the composition of the air is the same as on land with a mole fraction of N₂ being 0.78 (from the given standard composition of dry air), the partial pressure of N₂ is 0.78 x 2.71 atm = 2.1138 atm. Applying Henry's law:
Solubility = k × (partial pressure of N₂)
Solubility = 6.26 x 10⁻⁴ mol/(L·atm) × 2.1138 atm
Solubility = 1.323 x 10-3 mol/L
Therefore, the solubility of N₂ in the diver's blood at the given conditions is 1.323 x 10⁻³ mol/L.
A solution containing CaCl2 is mixed with a solution of Li2SO4 to form a solution that is 2.1 × 10-5 M in calcium ion and 4.75 × 10-5 M in sulfate ion. What will happen once these solutions are mixed? The Ksp for CaSO4 is 2.4 x 10-5.
Answer : The precipitate will not be formed when these solutions are mixed.
Explanation :
The chemical equation for the reaction of calcium chloride and lithium sulfate follows:
[tex]CaCl_2(aq)+Li_2SO_4(aq)\rightarrow 2LiCl(aq)+CaSO_4(s)[/tex]
We are given:
Concentration of calcium ion = [tex]2.1\times 10^{-5}M[/tex]
Concentration of sulfate ion = [tex]4.75\times 10^{-5}M[/tex]
[tex]K_{sp}=2.4\times 10^{-5}[/tex]
The salt produced is calcium sulfate.
The equation follows:
[tex]CaSO_4(s)\rightleftharpoons Ca^{2+}(aq)+SO_4^{2-}(aq)[/tex]
The expression of [tex]Q_{sp}[/tex] (ionic product) for above equation follows:
[tex]Q_{sp}=[Ca^{2+}]\times [SO_4^{2-}][/tex]
Putting values of the concentrations in above expression, we get:
[tex]Q_{sp}=(2.1\times 10^{-5})\times (4.75\times 10^{-5})\\\\Q_{sp}=9.9\times 10^{-10}[/tex]
There are 3 conditions:
When [tex]K_{sp}>Q_{sp}[/tex]; the reaction is product favored. (No precipitation)When [tex]K_{sp}<Q_{sp}[/tex]; the reaction is reactant favored. (Precipitation)When [tex]K_{sp}=Q_{sp}[/tex]; the reaction is in equilibrium. (Sparingly soluble)As, the [tex]K_{sp}>Q_{sp}[/tex]. The above reaction is product favored. This means that no salt or precipitate will be formed.
Hence, the precipitate will not be formed when these solutions are mixed.
Sodium tends lose a single electron in natural settings. Based on what you know, what are two other elements that tend to do the same thing?
Answer:
Lithium and Sodium
Explanation:
Losing and electron in natural setting is characterizes of elements in group one. These are elements known as the alkaline earth metals. They are the most electropositive elements on the periodic table.
These elements ionize by losing an electron yin their outermost shell to attain the configuration of the nearest noble gas. These elements are usually found in combined and rarely seen in uncombined state principally due to their very reactive nature.
Sodium naturally would ionize by losing one electron. Other elements capable of this even at a better rate because they are more electropositive are potassium and lithium. Both are also group one alkaline metals
If 75 gm's of codeine phosphate is dissolved in 1.5 liters of sterile water, what is the resultant percentage strength of the solution?
Answer:
w/v% = gm per 100 ml
75gm X gm 7500
------- = ------ == --------
1500ml 100ml 1500X
X=5%
Which of the following is true about a municipal bond with a put option? (A) An investor will exercise the option to put the bond if yields rise significantly. (B) An investor must have the issuer's permission to put the bond. (C) The market price of bonds is never affected by a put option feature. (D) Yields are usually higher for a new issue bond with a put option than for a new issue bond without a put option.
Answer:
(A) An investor will exercise the option to put the bond if yields rise significantly
Explanation:
A put option on the bond is a mechanism to allow the buyer of the bond the ability to compel the lender to repay the principal on the bond. The put option offers the buyer of the bond the ability to collect the principal of the bond anytime they choose until maturity for any purpose.
Recall that once the price drops (that is, the yield increases), put options are exercised. If the yield significantly increased, the put choice on a municipal bond is executed.
A sample of N2 gas is collected over water at 20o C and pressure of 1 atm. The volume collected is 250 Liters. What mass of N2 is collected?
Answer : The mass of nitrogen gas collected is, 290.9 grams
Explanation :
To calculate the mass of nitrogen gas we are using ideal gas equation:
[tex]PV=nRT\\\\PV=\frac{w}{M}RT[/tex]
where,
P = pressure of nitrogen gas = 1 atm
V = volume of nitrogen gas = 250 L
T = temperature of nitrogen gas = [tex]20^oC=273+20=293K[/tex]
R = gas constant = 0.0821 L.atm/mole.K
w = mass of nitrogen gas = ?
M = molar mass of nitrogen gas = 28 g/mole
Now put all the given values in the ideal gas equation, we get:
[tex](1atm)\times (250L)=\frac{w}{28g/mole}\times (0.0821L.atm/mole.K)\times (293K)[/tex]
[tex]w=290.9g[/tex]
Therefore, the mass of nitrogen gas collected is, 290.9 grams.
Given the unbalanced equation: N2(g) + H2(g) → NH3(g)
When the equation is balanced using the smallest whole-number coefficients, the ratio of moles of hydrogen consumed to moles of ammonia produced is
1 ) 1:3
2) 2:3
3) 3:1
4) 3:2
Answer:
Option-4 (3:2) is the correct answer.
Explanation:
Following steps are taken to balance the given unbalanced chemical equation.
Step 1: Write the unbalanced chemical equation,
N₂ + H₂ → NH₃
Step 2: Balance Nitrogen Atoms;
There are 2 nitrogen atoms on left hand side and 1 nitrogen atoms on right hand site therefore, to balance them multiply NH₃ on right hand side by 2 i.e.
N₂ + H₂ → 2 NH₃
Step 3: Balance Hydrogen Atoms;
Now, there are 2 hydrogen atoms on left hand side and 6 hydrogen atom on right hand site therefore, to balance them multiply H₂ on left hand side by 3 i.e.
N₂ + 3 H₂ → 2 NH₃
Now, the equation is balanced.
Step 4: Finding out mole ratios:
From balanced chemical equation it can be concluded that 3 moles of H₂ are involved in producing 2 moles of NH₃ hence, the mole ratio of consumption of H₂ to production of NH₃ is 3:2.
Answer:
4) 3:2
Explanation:
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The density of air under ordinary conditions at 25 degrees * C is 1.19g / L . How many kilograms of air are in a room that measures 9.0ft * 11.0ft and has a 10.0 ft ceiling?
Answer:
33.3 kg of air
Explanation:
This is a problem of conversion unit.
Density is mass / volume
Therefore we have to calculate the volume in the room, to be multiply by density. That answer will be the mass of air.
Volume of the room → 9 ft . 11 ft . 10 ft = 990 ft³
Density is in g/L, therefore we have to convert the ft³ to dm³ (1 dm³ = 1L)
990 ft³ . 28.3 dm³ / 1ft³ = 28017 dm³ → 28017 L
This is the volume of the room, if we replace it in the density formula we can know the mass of air in g.
1.19 g/L = Mass of air / 28017 L
Mass of air = 28017 L . 1.19 g/L → 33340 g of air
Finally, let's convert the mass in g to kg → 33340 g . 1kg / 1000 g = 33.3 kg
According to observations, the overall chemical composition of our solar system and other similar star systems is approximately (a) 98% hydrogen and helium, 2% all other elements combined; (b) 98% ice, 2% metal and rock; (c) 100% hydrogen and helium.
Answer:A
Explanation:
The solar system consist of the sun, the planets, stars and other objects. The chemical composition of the Sun consist mainly of Hydrogen and helium.
The sun is the largest object in the Solar system, it comprises nearly all the matter in the Solar System, Also the largest planet after the Sun are Jupiter and Saturn are giant planets forming almost the remaining matter of the solar system.
Like the Sun, the mass of Jupiter and Saturn are composed of roughly 98% hydrogen and helium with 2% of all the other elements combined.