Answer:
[tex]-3.896\times 10^{-12} N[/tex] is an electric force on the potassium ion due to the chloride ion.
Explanation:
Charge on potassium ion = [tex]q_1=1.602\times 10^{-19} C[/tex]
Charge on chlorine ion = [tex]=q_2=-1.602\times 10^{-19} C[/tex]
Separation between these two charges = r = [tex]7.700 nm=7.700\times 10^{-9} m[/tex]
[tex]1 nm=10^{-9} m[/tex]
Electric force on the potassium ion due to the chloride ion = F
Coulomb's law is given as ;
[tex]F=K\times \frac{q_1\times q_2}{r^2}[/tex]
[tex]q_1,q_2[/tex] = Charges on both charges
r = distance between the charges
K = Coulomb constant =[tex]9\times 10^{9} N m^2/C^2[/tex]
[tex]F=9\times 10^{9} N m^2/C^2\times \frac{1.602\times 10^{-19} C\times (-1.602\times 10^{-19} C)}{(7.700\times 10^{-9} m)^2}[/tex]
[tex]F=-3.896\times 10^{-12} N[/tex]
(negative sign indicates that attractive force is exerting between two ions)
[tex]-3.896\times 10^{-12} N[/tex] is an electric force on the potassium ion due to the chloride ion.
Final answer:
To find the electric force on a K+ ion due to a Cl- ion, Coulomb's law is used, employing the values for Coulomb's constant, the charge of each ion, and the distance between the ions.
Explanation:
The question refers to calculating the electric force on a K+ ion due to a Cl- ion across a membrane using Coulomb's law. Coulomb's law is defined as F = k * |q1*q2| / r^2, where F is the force between the charges, k is the Coulomb's constant (8.987 x 10^9 N m^2/C^2), q1 and q2 are the magnitudes of the charges (for K+ and Cl-, it's 1.602 x 10^-19 C), and r is the distance between the charges (7.700 nm or 7.700 x 10^-9 m). Plugging these values into Coulomb's law:
F = (8.987 x 10^9) * (1.602 x 10^-19)^2 / (7.700 x 10^-9)^2
After calculating, you will find the electric force exerted on the K+ ion due to the Cl- ion. This principle is crucial in understanding the movement of ions across cell membranes, influencing cell behavior and function.
A gas that has a volume of 28 L, a temperature of 45 °C, and an unknown pressure has its volume increased to 34 L and its temperature decreased to 35 °C. If the pressure is measured after the change to be 2.0 atm, what was the original pressure of the gas?A.1.6 atmB.2.5 atmC.3.2 atmD.4.1 atm
Answer:
The answer to your question is letter B. 2.5 atm
Explanation:
Data
P1 = ? P2 = 2 atm
T1 = 45°C = 318°K T2 = 35°C = 308°K
V1 = 28 L V2 = 34 L
Formula
Combined gas law
P1V1/T1 = P2V2/T2
solve for P1
[tex]P1 = \frac{T1P2V2}{V1T2}[/tex]
Substitution
P1 = (318 x 2 x 34) / (28 x 308)
Simplification
P1 = 21624 / 8624
Result
P1 = 2.5 atm
Silica, sio2, is formed on silicon as an electrically insulating layer for microelectronic devices. silica is formed when silicon is exposed to o2 gas at an elevated temperature. at 900˚c, it takes 90 minutes for the oxygen to diffuse from the surface to form a 0.06 micron (0.06 x 10-6 m) thick layer of sio2 on
Final answer:
Silicon dioxide, or silica, is integral to microelectronics as an electrical insulator and is used in applications such as isolation, gate insulation, and dopant diffusion. Chemical Vapor Deposition is commonly used to deposit silica thin films. The material's tetrahedral structure ensures stability under rapid temperature changes, vital for semiconductor industries.
Explanation:
Silicon dioxide (SiO2), often referred to as silica, plays a crucial role in the world of microelectronics. Its excellent electrical insulating properties make it essential for various applications within semiconductor devices. Silicon dioxide is employed for the isolation of conductive layers as well as for its dielectric properties which are used in gate insulation. Moreover, it serves to facilitate the diffusion of dopants from oxides and as a means to prevent the loss of dopants when capping films.
The Chemical Vapor Deposition (CVD) method is widely used for depositing thin layers of SiO2 during semiconductor processing. This is due to the unique challenges associated with its application in creating insulating thin films. Additionally, silica's unique diamond-like network structure allows for rapid temperature changes, making it invaluable in the steel, electronic, and semiconductor industries.
Silica's three-dimensional tetrahedral structure, where silicon atoms are bonded to oxygen, confers stability and resilience that is crucial for the high temperature processes involved in microelectronic device fabrication. Different forms of silicon dioxide, such as quartz and fused silica, contribute to the diversity of silica's properties and applications. Its naturally abundant presence and the numerous crystalline forms it can take, underscore the material's importance to technology and industry.
If x is a string, then x = new String("OH"); and x = "OH"; will accomplish the same thing. Group of answer choices True False
Answer:
True is the correct answer to the above question.
Explanation:
If x is a string then it can be assigned by the help of two ways in java:By the help of constructor:- When we write " x = new String("OH");", then it will create a pass a string "OH" into the constructor. It is because the String is a class in java and x is an object created by the constructor of the String class.With the help of assigning: The "x= OH", which assigns the value of x which is an object of String class it can also use the constructor to initialize the "OH" string on the class.The above question states the two scenarios which are defined above. Hence the question statement is true.Answer:
"True" is the correct answer to this question.
Explanation:
The program to the given question as follows:
Program:
public class data //defining class
{
public static void main (String [] aw)//defining the main method
{
String x="OH"; //defining string variable x and assign value
System.out.println("assign value: "+x); //print value
x = new String("OH"); //defining instance variable and assign value
System.out.println("assign value by creating instance: "+x); //print value
}
}
Output:
assign value: OH
assign value by creating instance: OH
Explanation of the program:
In the above java program, a class data is defined, inside the class the main method is declared, In this main method a string variable "x" is defined that holds a value "OH", then we the print function to print this variable value.
In the next line, An instance of variable x is created, which holds a value "OH" in its parameter. In this question, both are correct because both hold the same value.
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
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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Which of the following elements form cations (positively charged ions) readily? C, O, Na, Fe, As, Br, K
a. C, O, Na, Fe, As, Br, K
b. C, O, Na
c. Fe, As, Br, K
d. O, Na, Fe
e. Na, Fe, K
Answer:
e. Na, Fe, K
Explanation:
The group of elements that will be positive charge ions is metal. You can find metal in the first 2 columns of the periodic table and in the transition area. Natrium/sodium (Na), iron (Fe), and kalium/potassium(K) categorized as metal and they will form positive charge ions.
On the other hand carbon(C), oxygen (O), arsenic(As) and bromine(Br) is gas and will form negative charge ions. Gas located on the right side of the periodic table.
The elements that readily form positively charged ions, or cations, are more often metals like Sodium (Na), Iron (Fe), and Potassium (K). Therefore, the correct answer from the options given is 'Na, Fe, K'. option e.
Explanation:The elements that form cations, or positively charged ions, readily are elements that tend to lose electrons. This characteristic is typically associated with metals. In the options given, Na (Sodium), Fe (Iron), and K (Potassium) are the ones which are more likely to form cations. So, the correct answer to your question: 'Which of the following elements form cations (positively charged ions) readily: C, O, Na, Fe, As, Br, K?' would be option e. Na, Fe, K.
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Find the molecular formula of a compound that contains 42.56 g of palladium and 0.80 g of hydrogen. The molar mass of the compound is 216.8 g/mol.
Answer:
The answer to your question is Pd₂H₄
Explanation:
Data
mass of palladium = 42.56 g
mass of hydrogen = 0.8 g
Process
1.- Convert the grams of each substance to moles
106 g of Pd ----------------- 1 mol
42.56 g of Pd ------------- x
x = (42.56 x 1)/106
x = 0.402 moles
1 g of H --------------------- 1 mol
0.8 g of H ------------------ x
x = (0.8 x 1)/1
x = 0.8 moles
2.- Divide by the lowest number of moles
Palladium = 0.402/0.402 = 1
Hydrogen = 0.8/0.402 = 1.99 ≈ 2
3.- Write the empirical formula
PdH₂
4.- Calculate the mass of the empirical formula
PdH₂ = 106 + 2 = 108
5.- Divide the molar mass by the molar mass of the empirical formula
216.8/108 = 2
6.- Get the molecular formula
2(PdH₂) = Pd₂H₄
The molecular formula of the compound is determined to be Pd2H4, based on the calculated mole ratio of palladium to hydrogen and considering the given molar mass of the compound.
Explanation:To find the molecular formula of a compound, we first determine the mole ratio between palladium and hydrogen. The molar mass of palladium (Pd) is 106.42 g/mol and the molar mass of hydrogen (H) is 1.008 g/mol. Therefore, 42.56 g of Pd is equivalent to 42.56/106.42 = 0.4 mol, and 0.80 g of H is equivalent to 0.80/1.008 = 0.793 mol. This simplifies to a ratio of 1:2, indicating the empirical formula of the compound is PdH2.
Next, we compare the molar mass of the empirical formula with the molar mass given for the compound. The molar mass of PdH2 is 1.008*2 + 106.42 = 108.436 g/mol. The molar mass given for the compound is 216.8 g/mol. Therefore, the molecular formula is twice the empirical formula, result to be Pd2H4.
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The best mixture of antifreeze and water is 50% antifreeze, 50% water. The cooling system in your car has a mixture of 6.00L water and 6.00 L ethylene glycol (antifreeze). The molality of the solution is 17.9m. The chemical formula of antifreeze is C2H6O2 and its density is 1.1132 g/cm3.
If the summer temperatures rise and the coolant reaches a temperature of 108°C, will it boil?
No, it would boil at 109.13°C
No, it would boil at 123.09°C
No, it would boil at 119.13°C
Yes, it would boil at 99.0°C
Yes, it would boil at 100.13°C
Answer:
The correct answer is No, it would boil at 109.13°C
Explanation:
This question can be solved by knowing the boiling point elevation formula and the fact that ethylene glycol dissolves in water without dissociation
The boiling point elevation formula is given by
ΔT = i × [tex]K_{b}[/tex] ×[tex]m_{solute}[/tex]
Where [tex]K_{b}[/tex] = 0.51 °C/mole
i = Vant't Hoff factor
m = molality of the solution
When ethylene glycol, C2H6O2, (antifreeze) enters into solution in water it disociates into
C2H6O2 (aq) ---> 2OH(-1)(aq) + C2H4(+2)(aq)
Thus one mole of C2H6O2 dissociates into two moles of hydroxyl ions and one mole of C2H4(+2) ion
Hence the Van't Hoff factor, i, = 3
Therefore the mass of the mole
Therefore ΔT = 3 × 0.51 × 17.9 = 27.387 K
However Ethylene formula = (CH2OH)2 it dissolves in water without dissociation
Therefore i = 1
and ΔT = 1 × 0.51 × 17.9 = 9.129 ≅ 9.13
Hence at the boiling point of the water with antifreeze dissolved in it it
Boiling point of water + Boiling point elevation = 100 + 9.13 = 109.13 °C
The water will not boil until it reaches 109.13 °C
The coolant mixture will not boil at 108°C. The calculated boiling point elevation suggests it will boil at approximately 109.16°C, based on the molality and the boiling point elevation constant for water.
Explanation:To determine whether the coolant mixture in your car will boil at 108°C, we can use the concept of boiling point elevation. The boiling point of a solution increases when a solute is added to a solvent due to the colligative properties of the solution. Using the molality provided (17.9m), and knowing the boiling point elevation constant (Kb) for water is approximately 0.512 °C/m, we can calculate the boiling point elevation.
ΔTb = i * Kb * m
Where ΔTb is the boiling point elevation, i is the van't Hoff factor (i = 1 for ethylene glycol as it does not dissociate in solution), Kb is the ebullioscopic constant for water, and m is the molality of the solution.
ΔTb = 1 * 0.512 °C/m * 17.9m = 9.16 °C
The normal boiling point of water is 100°C, so adding the elevation to the normal boiling point gives us:
100°C + 9.16°C = 109.16°C
Therefore, the solution will not boil at 108°C because it would boil at approximately 109.16°C.
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.
If 0.500 moles of sulfuric acid and 0.500 moles of Aluminum hydroxide react to make water and aluminum sulfate. a) which reactant is the limiting reactant b) How many moles of Aluminum sulfate is produced?
Answer:
The correct answer to a) is sulfuric acid
b) 1/6 moles of aluminum sulfate is produced
Explanation:
To solve this we need to write out the balaned chemical equation as follows
Al(OH)3(s) + 3 H2SO4(aq) -----> Al2 (SO4)3(aq) + 6 H2O(l)
Here we see that one mole of aluminium hydroxide reacts with three moles of sulphuric acid to form one mole of aluminium sulphate and six ,oles of water
Hence the limiting reactant in this question is the sulphuric acid as 0.5 moles of alumininium hydroxide requires 1.5 moles of sulphuric acid to completely use up the aluminium hydroxide present
Dividing the number of moles of the reactants present by the amount of moles of sulphuric acid required we have
Hence 3 mole of sulphric acid combines with 1 mole of Aluminium hydroxide
0.5 mole of sulfuric acid combines with 0.5÷3 or 1/6 mole of Aluminium hydroxide
Hence the correct answer is sulfuric acid
b) to solve this, since three moles of sulfuric acid produces one mole of aluminium sulfate then 0.5 moles of sulfuric acid produces 0.5/3 or 0.166 mole of aluminum sulfate
hence 1/6 moles of aluminum sulfate is produced
A reducing agent gets oxidized as it reacts. A reducing agent gets oxidized as it reacts. true false
A reducing agent does get oxidized as it reacts in a process known as a redox reaction. During this reaction, the reducing agent loses electrons, effectively donating them to another substance. Disproportion reactions are also possible, where the same substance gets oxidized and reduced.
Explanation:Yes, the statement is true: A reducing agent gets oxidized as it reacts. In redox reactions, the substance that is oxidized loses electrons and is therefore referred to as the reducing agent. For example, in the reaction, 'Al(s) + NiO(s) --> Al2O3(s) + Ni(s)', aluminum (Al) is the reducing agent as it gets oxidized from 0 to +3 oxidation state while reducing nickel oxide (NiO).
Redox reactions involve the transferring of electrons from one atom (which gets oxidized) to another (which gets reduced). In the process, the reducing agent is oxidized because it essentially donates its own electron to the other substance.
There are also cases of disproportion reactions where the same substance gets oxidized and reduced. These reactions are very interesting in the context of oxidation-reduction chemistry.
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By what factor does the rate change in each of the following cases (assuming constant temperature)? (a) A reaction is first order in reactant A, and [A] is doubled. (b) A reaction is second order in reactant B, and [B] is halved. (c) A reaction is second order in reactant C, and [C] is tripled.
Answer:
a) 2
b) 1/4
c) 9
Explanation:
a) for a first order reaction in reactant A
r initial = k*[A initial]
then if the concentration is doubled [A final ]= 2*[A initial] , then
r final = k*[A final ] = 2* k*[A initial] = 2*r initial
then the velocity changes by a factor of 2
b) for a second order reaction in reactant B
r initial = k*[B initial]²
then if the concentration is halved: [B final ]= [B initial]/2 , then
r final = k*[B final ]² = k*( [B initial]/2 )² =k* [B initial]² /4 = r initial /4
then the velocity changes by a factor of 1/4
c) for a second order reaction in reactant C
r initial = k*[C initial]²
then if the concentration is tripled : [C final ]= 3* [C initial] , then
r final = k*[C final ]² = k*( 3*[C initial] )² =k* [C initial]² *9 = 9 * r initial
then the velocity changes by a factor of 9
For first-order reactions, doubling the concentration of the reactant doubles the reaction rate. For second-order reactions, halving the concentration reduces the rate by a factor of four, while tripling the concentration increases the rate by a factor of nine.
Explanation:(a) For a reaction that is first order in reactant A, if [A] is doubled, the rate of the reaction will also double. This is because the rate of a first-order reaction is directly proportional to the concentration of the reactant. When [A] is doubled, the rate is doubled as well.
(b) For a reaction that is second order in reactant B, if [B] is halved, the rate of the reaction will decrease by a factor of four. This is because the rate of a second-order reaction is directly proportional to the square of the concentration of the reactant. When [B] is halved, the concentration is squared and the rate is decreased by a factor of four.
(c) For a reaction that is second order in reactant C, if [C] is tripled, the rate of the reaction will increase by a factor of nine. Similar to the previous case, the rate of a second-order reaction is directly proportional to the square of the concentration of the reactant. When [C] is tripled, the concentration is squared and the rate is increased by a factor of nine.
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The density of water is 1.00 gram/milliliter. What is the volume in milliliters of 1.00 mole of water? Express your answer to the correct number of significant figures.
Answer:
18.00 mL is the volume in mL of 1 mol of water
Explanation:
Water density = water mass / water volume
1 g/mL = water mass / water volume
1 mol of water weighs 18 g. Therefore, 1 g/mL = 18 g / water volume
water volume = 18 mL
Answer:
18.0
Explanation:
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For the procedural error, indicate if the error will affect the actual yield of copper(II) saccharinate product and if it does, will it raise or lower the actual yield:__________
Washing the crystals with hot water
Answer:
Lowers the actual yield
Explanation:
Aluminum reacts with excess hydrochloric acid to form aqueous aluminum chloride and 190.6 ml of hydrogen gas (collected over water at 26°C and 0.89 atm). (Vapor pressure of water at 26ºC = 25.2 mmHg.) How many grams of aluminum reacted? Enter to 4 decimal places.
Answer:0.119g
Explanation:equation of rxn is
2Al+6HCl=2AlCl3+3H2
From ideal gas eqn
PV=nRT
n=PV/RT
P here is the partial pressure of H2 from the qtn.According to Dalton law of partial pressure, PT=PH2+PH20
PT=0.89atm given
PH20=25.2mmhg given=25.2/760atm,=0.033atm
PH2=PT-PH20
PH2=0.89-0.033=0.857atm
T=26+273=299K
R=0.082atmdm^-3mol^-1K^-1
V=190.6ml=190.6cm3=190.6/1000=0.1906dm3
n=PV/RT
n=0.857*0.1906/0.082*299
=0.00667moles of H2.
From the eqn of reaction,
2moles of Al reacts to gv 3moles of H2
xmoles of Al will give 0.00667moles of H2
xmoles=0.00667*2/3 (cross multiplying)=0.00444moles of Al
From the relationship, n=mass/MW
mass=MW*n
MW of Al=27g/mol
mass=0.0044moles*27g/moles
mass=0.119grams of Al.
The mass of aluminum that has reacted is 0.119 g. The mass of the reactant can be calculated by finding its moles.
How to calculate the mass of the reactant?the mass of the reactant can be calculated by finding its moles in the reaction and putting the value in the mole formula.
The given reaction is:
[tex]\bold {2Al+6HCl\rightarrow 2AlCl_3+3H_2}[/tex]
First, calculate the moles of the Hydrogen from the ideal gas equation,
[tex]n=\dfrac {0.857\times 0.1906}{0.082\times 299}\\\\ n = \rm 0.00667 \ moles \ of \ H2.[/tex]
The molar ratio between Al and [tex]\bold {H_2 }[/tex] is 3:2.
Thus, moles of Aluminium is:
[tex]\text{Moles of Al} = 0.00667\times \dfrac 23 \\\\\text{Moles of Al} = 0.00444 \rm \ moles[/tex]
Thus the mass of Aluminium,
[tex]m ={\rm 0.0044 \ moles\times 27 \ g/moles}\\\\m = 0.119\rm \ g[/tex]
Therefore, the mass of aluminum that has reacted is 0.119 g.
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The following statements about blood are true except that __________.
Except for __________, the following statements about blood are true.
a.The viscosity is three to five times greater than water.
b The pH is slightly acidic.
c. It contains about 55% plasma.
D. It contains dissolved gases.
Answer:Except for the pH is slightly acidic, the following statements about blood are true.
Explanation:Viscosity is an inherent feature of liquid compared to the inner resistance of nearby fluid films pushing through one another. All blood has a much greater viscosity than water. Authorities regard these matters as “non-Newtonian fluids,” of which ketchup and blood are excellent illustrations.
Blood is usually slightly basic, with a typical pH scale of approximately 7.35 to 7.45. Plasma composes 55% of the whole blood volume. The albumin included in plasma restricts the blood from dropping too much water. The plasma of vertebrates also includes dissolved gases.
Blood is a fluid connective tissue that is 92 percent water. It is slightly more acidic, viscous, and salty than water.
Explanation:Blood is a fluid connective tissue composed of plasma, dissolved substances, and blood cells. It is about 92 percent water, making the statement provided true. However, blood is slightly more acidic than water, slightly more viscous than water, and slightly more salty than seawater, making these statements false. Red blood cells carry oxygen, white blood cells defend the body, and platelets help blood clot.
Crime scene investigators keep a wide variety of compounds on hand to help with identifying unknown substances they find in the course of their duties. One such investigator, while reorganizing their shelves, has mixed up several small vials and is unsure about the identity of a certain powder. Elemental analysis of the compound reveals that it is 63.57% carbon, 6.000% hydrogen, 9.267% nitrogen, 21.17% oxygen by mass. Which of the following compounds could the powder be?
a.) C11H15NO2 = 3,4-methylenedioxymethamphetamine (MDMA), illicit drug
b.) C3H6NO3 = hexamethylene triperoxide diamine (HMTD), commonly used explosive
c.) C21H23NO5 = heroin, illicit drug
d.) C8H9NO2 = acetaminophen, analgesic
e.) C7H5N3O6 = 2,4,6-trinitrotoluene (TNT), common used explosive
f.) C17H19NO3 = morphine, analgesic
g.)C10H15N = methamphetamine, stimulant
h.) C4H5N2O = caffeine, stimulant
Answer:
Option d: C₈H₉NO₂ = acetaminophen, analgesic
Explanation:
% composition of compound is:
63.57 g of C
6 g of H
9.267 g of N
21.17 g of O
First of all we divide each by the molar mass of the element
63.57 g / 12 gmol = 5.29 mol of C
6 g of H / 1 g/mol = 6 mol H
9.267 g of N / 14 g/mol = 0.662 mol of N
21.17 g of O / 16 g/mol = 1.32 mol of O
We divide each by the lowest value, in this case 0.662
5.29 / 0.662 = 8
6 / 0.662 = 9
0.662 / 0.662 = 1
1.32 / 0.662 = 2
Molecular formula of the compound is C₈H₉NO₂
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.
When a chemical reaction is in equilibrium,
a. the reaction is proceeding at its maximum rate.
b. there is no net change in the amount of substrates or products.
c. the reaction has stopped.
d. there are equivalent amounts of substrates and products.
Answer: b. there is no net change in the amount of substrates or products.
Explanation:
The reactions which do not go on completion and in which the reactant forms product and the products goes back to the reactants simultaneously are known as equilibrium reactions.
Equilibrium state is the state when reactants and products are present but the concentrations does not change with time.
[tex]A\rightleftharpoons B[/tex]
For a chemical equilibrium reaction, equilibrium state is achieved when the rate of forward reaction becomes equals to rate of the backward reaction.
The following peptides are subjected to normal electrophoretic analysis at pH 6.0. State whether the peptides will migrate towards the cathode or anode and predict the relative rate of migration of each peptide. a.GlyArg Phe.b.Gly.Arg Phe.c.Glu.Glu Phe.d.GIy.Glu
Answer:
The peptide will definitely migrated towards cathode (negative terminal)
Explanation:
The positively and negatively charged side chains of proteins cause them to behave like amino acids in an electrical field; that is, they migrate during electrophoresis at low pH values to the cathode (negative terminal) and at high pH values to the anode (positive terminal). The isoelectric point, the pH value at which the protein molecule does not migrate, is in the range of pH 5 to 7 for many proteins.
Ice Station Bravo near the North Pole launched a helium-filled balloon to check atmospheric conditions. At sea level (1.0 atm) where the balloon was launched, it had a volume of 0.93m^3 . It rose to an altitude of 18000m where the atmospheric pressure dropped to 0.072atm.
What is the volume of the balloon at that altitude assuming that the temperature was the same at sea level?
Answer:
12.9 m³ is the new volume
Explanation:
As the temperature keeps on constant, and the moles of the gas remains constant too, if we decrease the pressure, the volume will increase. If the volume is decreased, pressure will be higher.
The relation is this: P₁ . V₁ = P₂ . V₂
1 atm . 0.93m³ = 0.072 atm . V₂
0.93m³ .atm / 0.072 atm = V₂
V₂ = 12.9 m³
In conclusion and as we said, pressure has highly decreased so volume has highly increased.
If the density of a 21.71 m (molal) solution of ethanol, C2H5OH, in water is 0.914 g/mL, what is the molarity of ethanol in this solution?
The molarity of the given ethanol solution can be computed using its given molality and the density of the solution. The molality tells us the moles of ethanol per kilogram of water, and the density allows us to convert this to moles per liter, generally yielding a larger molarity for the same solution. Thus, the molarity of ethanol in this solution would be 914 M.
Explanation:To calculate the molarity of the ethanol solution, we first need to know that the definition of molality (m) is the moles of solute (in this case, ethanol) per kilogram of solvent (here, water). Given that the density of the solution, is 0.914 g/mL, we can convert this to kg/L for ease of calculation. So, 0.914 g/mL is equivalent to 914 kg/m^3.
Next, let's take 21.71 molal solution, implying there are 21.71 moles ethanol in 1 kilogram of water. Since the density of the solution is 0.914 g/mL (or 914 kg/m^3), there would be 914 moles of ethanol in 1 m^3 of solution (since 1 L = 1 m^3).
Thus, the molarity (M), which is defined as the moles of solute per liter of solution, of the ethanol in the solution would be 914 M.
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if you take a dried out white rose and place the stem in food coloring, will the rose turn the color you put the stem in
Answer:
Yes, this is true. The reason is that the flower transpires and sucks the water in and distributes it as much as it can. You can also flip it upside down and hang it with petals down , allowing the liquid to enter the flower and then retaining color for longer periods of time and having more color.
Explanation:
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
When a current is passed through a solution of salt water, sodium chloride decomposes according to the following reaction: NaCl + H2O → NaOH + Cl2 + H2 Balance the equation. Choose "blank" if no coefficient is needed. NaCl + H2O → NaOH + Cl2 ++ H2
Answer:
The answer to your question is 2NaCl + 2H₂O ⇒ 2NaOH + Cl₂ + H₂
Explanation:
Original chemical equation
NaCl + H₂O ⇒ NaOH + Cl₂ + H₂
Reactant Element Products
1 Na 1
1 Cl 2
2 H 3
1 O 1
This reactions is unbalanced
2NaCl + 2H₂O ⇒ 2NaOH + Cl₂ + H₂
Reactant Element Products
2 Na 1
2 Cl 2
4 H 4
2 O 2
Now, the reaction is balanced
Answer:
NaCl + 2 H2 O → 2 NaOH + Blank Cl 2 + Blank H 2
describe how these nutrients and elements such as carbon, oxygen, and nitrogen are cycled through ecosystems.
Answer:
Carbon Cycle
Steps of the Carbon Cycle
CO2 is removed from the atmosphere by photosynthetic organisms (plants, cyanobacteria, etc.) and used to generate organic molecules and build biological mass. Animals consume the photosynthetic organisms and acquire the carbon stored within the producers. CO2 is returned to the atmosphere via respiration in all living organisms. Decomposers break down dead and decaying organic matter and release CO2. Some CO2 is returned to the atmosphere via the burning of organic matter (forest fires). CO2 trapped in rock or fossil fuels can be returned to the atmosphere via erosion, volcanic eruptions, or fossil fuel combustion.Nitrogen Cycle
Steps of the Nitrogen Cycle
Atmospheric nitrogen (N2) is converted to ammonia (NH3) by nitrogen-fixing bacteria in aquatic and soil environments. These organisms use nitrogen to synthesize the biological molecules they need to survive. NH3 is subsequently converted to nitrite and nitrate by bacteria known as nitrifying bacteria. Plants obtain nitrogen from the soil by absorbing ammonium (NH4-) and nitrate through their roots. Nitrate and ammonium are used to produce organic compounds. Nitrogen in its organic form is obtained by animals when they consume plants or animals. Decomposers return NH3 to the soil by decomposing solid waste and dead or decaying matter. Nitrifying bacteria convert NH3 to nitrite and nitrate. Denitrifying bacteria convert nitrite and nitrate to N2, releasing N2 back into the atmosphere.Oxygen Cycle
Oxygen is an element that is essential to biological organisms. The vast majority of atmospheric oxygen (O2) is derived from photosynthesis. Plants and other photosynthetic organisms use CO2, water, and light energy to produce glucose and O2. Glucose is used to synthesize organic molecules, while O2 is released into the atmosphere. Oxygen is removed from the atmosphere through decomposition processes and respiration in living organisms.
Explanation:
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.
A(n) _______ solution has a higher concentration of water and lower concentration of solute than the cell placed in the solution.
Answer : A hypotonic solution has a higher concentration of water and lower concentration of solute than the cell placed in the solution.
Explanation :
Solution : It is made up of the combination of amount solute and solvent.
Isotonic solutions : It is defined as the solutions in which the concentration of solute inside the cell and outside the cell is same.
Hypotonic solutions : It is defined as the solutions in which the concentration of solute inside the cell is lower than outside the cell.
For example : Diluted sugar syrup
Hypertonic solutions : It is defined as the solutions in which the concentration of solute inside the cell is higher than outside the cell.
For example : Concentrated sugar syrup
Hence, a hypotonic solution has a higher concentration of water and lower concentration of solute than the cell placed in the solution.
The term for a solution that has a higher concentration of water and a lower concentration of solute than a cell is 'hypotonic'. In this scenario, water moves into the cell via osmosis.
Explanation:A(n) hypotonic solution has a higher concentration of water and lower concentration of solute than the cell placed in the solution. In biology, we often talk about the relationship between cells and their surrounding environment in terms of tonicity. In a hypotonic environment, there is less solute (like salt or sugar) outside the cell compared to inside the cell. This causes water to move into the cell by osmosis, because water moves from areas of high concentration to areas of low concentration until equilibrium is reached.
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Which of the following is the correct ranking of the three bonds and interactions in order from highest to lowest in terms of their bond strength between two side chains of a protein in their tertiary structure?
I. Disulfide bond between two cystines
II. Hydrophobic interactions between two leucines
III. H-bonding in water
Answer:
I > III > II
Explanation:
I) A disulfide bond between two cystines is created when a sulfur atom from one cystine forms a strong, single covalent bond with a sulfur atom from a second cystine. When a disulfide bond is created, each cystine loses one hydrogen atom. The atom count is 11 for a cystine in mid-chain, but changes to 10 if the cystine joins with another in a disulfide bond. This lead to a much more stable intermolecular interaction.
III) Hydrogen Bonding in water
These hydrogen bonds are at best an interaction, inducing slight positive and negative charges in the Hydrogen and Oxygen/Nitrogen atoms.
The Hydrophilic amino acids have O & N atoms, which form hydrogen bonds with water. These atoms have an uneven distribution of electrons, creating a polar molecule that can interact and form hydrogen bonds with water.
The hydrogen bonds aren't as strong as the covalent bonds in disulfides.
II) Hydrophobic interactions between two leucines
A hydrophobic interaction is formed between two nonpolar molecules.
It describes the preference of nonpolar molecular surfaces to interact with other nonpolar molecular surfaces, thereby displacing water molecules from the interacting surfaces.
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.