Consider the chemical reaction described by the following equation.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
ΔH = −58 kJ
What would be the enthalpy, in kilojoules, for the reaction
2NaCl(aq) + 2H2O(l) → 2HCl(aq) + 2NaOH(aq)
ΔH = ?

Answer : The pressure of the helium gas inside the container would be, 32.1 atm

Explanation :

To calculate the pressure of the gas we are using ideal gas equation as:

[tex]PV=nRT[/tex]

where,

P = Pressure of [tex]He[/tex] gas = ?

V = Volume of [tex]He[/tex] gas = 7.5 L

n = number of moles [tex]He[/tex] = 10 mole

R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]

T = Temperature of [tex]He[/tex] gas = [tex]20^oC=273+20=293K[/tex]

Putting values in above equation, we get:

[tex]P\times 7.5L=10mole\times (0.0821L.atm/mol.K)\times 293K[/tex]

[tex]P=32.1atm[/tex]

Thus, the pressure of the helium gas inside the container would be, 32.1 atm

Answer: The proteins were no longer soluble in the blood.

Answer:

True

Explanation:

This are acts or actions that concur with situational expectations.

No I don't think it works that way. What you could do is drink one energy drink, wait 5 hours and then drink another.

Answer:

10 hrs of energy

Explanation:

these bottles hold energy for 5hrs so if it is double you have double the time of energy

Answer:

The right answer to this question is no, all of the liquid will not evaporate, there will be 8.61 ×10⁻⁴ moles or 6.22×10⁻² grams left in the container

At equilibrium the pressure in the container will be the vapor pressure of liquid pentane which is = 100. mm Hg

Explanation:

To solve this we list the known values as follows

vapor pressure of liquid pentane = 100 mmHg = 13.33 KPa

Temperature T = 260 K

Volume of container = 350 mL = 0.00035 m³

The number of moles of liquid pentane = n

The universal gas constant = R = 8.314 J/(mol·K)

Thus From the ideal gas equation PV = nRT →

Thus plugging in the values in the above equateion we have

n = [tex]\frac{PV}{RT} = \frac{(13330)(0.00035)}{(8.314)(260)}[/tex] = 2.16×10⁻³ moles

Hence the number of moles in 0.218 g sample of liquid   pentane C₅H₁₂ with molar mass = 72.15 g/mol = 0.218/72.15 = 3.02×10⁻³ moles

Hence the number of moles present in the sample placed in the closed evacuated container = 3.02×10⁻³ moles

However number of moles to completely evaporate at 100 mmHg and 260 K is 2.16×10⁻³ moles hence, 3.02×10⁻³ moles - 2.16×10⁻³ moles,  or 8.61 ×10⁻⁴ moles will be left in the container

converting the value in moles to mass we have number of moles, n = mass/(molar mass)

Therefore the mass = number of moles × molar mass = 8.61 ×10⁻⁴ × 72.15 = 6.22 × 10⁻² grams left in the container

The pressure in the container at equilibrium will be vapor pressure of liquid pentane C₅H₁₂, or 100. mm Hg

In the described closed system, all of the liquid pentane will evaporate due to the vapor pressure. The final pressure when equilibrium is reached will be the vapor pressure, which is 100.0 mm Hg.

The subject of the question is the vapor pressure of pentane, C5H12, which is a concept from Chemistry.

The given vapor pressure is 100.0 mm Hg at 260 K. In this closed system the liquid and vapor will come to equilibrium at the given temperature, at which point the vapor pressure will be equal to the given 100.0 mm Hg.

The sample mass of 0.218 g doesn't exceed the amount needed for evaporation at the equilibrium pressure. Therefore, all of the liquid pentane will evaporate.

The final pressure in the container when equilibrium is reached will be the vapor pressure, which is 100.0 mm Hg.

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Answer:

The molarity is 1.33×10^-7M

Explanation:

15ppb = 15g of chlorobenzene/10^9g of solution × 1 mole of chlorobenzene/112.6g × 1g of solution/mL = 1.33×10^-10mol/mL × 1000mL/1L = 1.33×10^-7mol/L

The molarity of a chlorobenzene solution with a concentration of 15 ppb is calculated by first converting the ppb measurement to mass per volume and then using the molar mass to find the number of moles per liter, resulting in a molarity of 1.33 × 10⁻⁷ M.

To calculate the molarity of chlorobenzene in a solution with a concentration of 15 ppb, we first need to convert this concentration into a mass-volume relationship. Since 1 ppb is equivalent to 1 µg/L, 15 ppb is equal to 15 µg/L, which is the same as 0.015 mg/L.

Using the molar mass of chlorobenzene, which is 112.6 g/mol, we can convert this mass into moles using the following equation:

The solution has a concentration of 15 µg/L, which is equal to 0.015 mg/L or 0.000015 g/L. So the amount of moles of chlorobenzene is:

Number of moles = 0.000015 g / 112.6 g/mol = 1.33 × 10-7 moles/L

Therefore, the molarity of the chlorobenzene solution is 1.33 × 10-7 M.

Sound waves are D) Mechanical and Longitudinal

Explanation:

In physics, waves are classified into two types:

Moreover, waves are further classified into:

Sound waves are oscillations of a medium that occurs back-and forth along the direction of propagation of the wave. Therefore, they are mechanical (they need a medium to propagate) and longitudinal. So the correct answer is

D) Mechanical and Longitudinal

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Final answer:

Sound waves are best described as Mechanical and Longitudinal waves, as they need a medium to travel and have disturbances parallel to the direction of propagation in fluids.

Explanation:

The set of terms that best describes sound waves is Mechanical and Longitudinal. Sound waves require a medium to travel through, which makes them mechanical waves. Furthermore, in fluids such as air and water, sound waves are longitudinal because their disturbances (variations in pressure) are parallel to the direction of propagation. In solids, sound waves can be both longitudinal and transverse; however, in the context of your question which seems to imply a generic scenario, the focus is often on sound in fluids.

Answer:

Critical Point

Explanation:

You can no longer distinguish a liquid from a gas when a object hits it's critical point.

Answer:

the velocity of the tomato will be  v  = u + gt

kinetic energy = 9.64 h

Explanation:

At that time, let the initial speed be u.

The acceleration due to gravity be g

The final velocity will be given as v = u + at

but a = g = 9.81 m/s² [acceleration due to gravity]

so the final velocity will be given as v = u + at

Let the potential energy be Ep

Before landing the ground, 93.8 % of the potential energy will be converted to kinetic energy. Therefore, the calculation will be as follows:

Ep = mgh

Kinetic energy Ek = 1/2mv²

But, Ek = 0.938 Ep

            = 0.983 × gh

            = 9.64 h

where h is the height of the object from the ground.

Answer:

calculate the molecular formula of a compound with the empirical formula CH2O and a molar mass of 150g/mol

the molecular formula is [tex]C_{5} H_{10}O_{5}[/tex]

The molecular formula of a compound is the formula comprising of the constituent elements chemical symbols each of which carries the number of atoms of that element present in a molecule of the compound appearing in the smallest whole number ratio to other eatoms present in the compound

Explanation:

The masses of the constitent element is determined forst from which the number of moles is then calculated by dividing the mass by the molar mass then then each calculated molar mass value is divided by the smallest  number of moles calculated from the previous step so the value calculated is then rounded up to the nearest whole number giving the ratios of the moles of the elements in the compound which represents the subscripts in the empirical formula of the compound.

If the subscrips are in fractions, then multiply each of them by the same number to derive the smallest whole number factor, that is if the calculated formula contains a facor of 0.5, multiply by 2

Mass pf Carbon  = 12g

mass of Hydrogen = 1g

molar mass of oxygen - 16g

Total mass of CH2O = 30g

Therefore dividing molar mass by empirical formula mass = 150g/30g = 5

Hence our factor is 5

multiplying each subscript of the empirical formula by 5 gives

C5H10O5 hence the molecular formula is [tex]C_{5} H_{10}O_{5}[/tex]

Answer:

Molecular formula of the compound is C₃H₆O

Explanation:

Firstly let's determine the moles of the vapor (gas) with the Ideal Gases Law, so it can give us the molar mass with the mass and afterwards we can work with the percent composition.

Pressure . Volume = moles . Ideal Constant Gases . Temperature in K

Temperature in K = T°C + 273 → 150°C + 273 = 423K

P . V = n . R . T

n = (P .V) / (R. T)

n = 1 atm . 0.5L / (0.082 . 423K)

n =  0.0144 moles

These are the moles for 0.8365 g, so let's determine the molar mass

Molar mass (g/mol) = 0.8365 g / 0.0144 mol → 58.02 g/mol

Percent composition means:

100 g of compound have 62.04 g of C

100 g of compound have 10.41 g of H

100 g of compound have 27.54 g of O

Let's make the rule of three:

100 g of compound have __ 62.04 g of C __ 10.41 g of H __ 27.54 g of O

The 58.02 g of compound must have:

(58.02 g . 62.04 g) / 100 g  = 36 g of C

(58.02 g . 10.41 g) / 100 g  = 6 g of H

(58.02 g . 27.54 g) / 100 g  = 16 g of O

Let's find out the moles of each

Mass / Molar mass

36 g / 12 g/mol = 3 C

6 g / 1 g/mol = 6 H

16g / 16 g/mol = 1 O

The molecular formula of the given compound is C₃H₆O. The molecular formula can be determined by finding the ratio of each element in the compound.

It can be determined by finding the ratio of each element in the compound.

First, calculate the moles of the compound from the ideal gas formula,

[tex]n = \dfrac {1 {\rm\ atm \times 0.5L} }{(0.082 \times 423{\rm \ K})}\\\\n = 0.0144 \rm \ moles[/tex]

Then calculate the molar mass of the compound,

[tex]m = {\rm \dfrac {0.8365 \ g }{0.0144 \ mol}} \\\\m = 58.02 \rm \ g/mol[/tex]

Then calculate the mass of individual elements in the 58.02 g of compound:

[tex]\text{ Mass of Carbon} = \dfrac {58.02 {\rm \ g} \times 62.04 {\rm \ g}}{100 {\rm \ g}}\\ \text{ Mass of Carbon} = 36 \rm \ g[/tex]

[tex]\text{ Mass of Hydrogen } = \dfrac {58.02 {\rm \ g} \times 10.14 {\rm \ g}}{100 {\rm \ g}}\\ \text{ Mass of Hydrogen } = 6 \rm \ g[/tex]

[tex]\text{ Mass of Oxygen } = \dfrac {58.02 {\rm \ g} \times 27.54 {\rm \ g}}{100 {\rm \ g}}\\ \text{ Mass of Oxygen } = 16 \rm \ g[/tex]

Now find the moles of each element, we get

3 moles of Carbon

6 moles of Hydrogen

1 mole of Oxygen

Therefore, the molecular formula of the given compound is C₃H₆O.

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Final answer:

To calculate the percent composition of aluminum and magnesium in the sample, we use stoichiometry. Moles of hydrogen gas produced help to establish moles of the metals reacting, but without complete reaction equations for the alloy, or more data, a definitive calculation cannot be provided.

Explanation:

To calculate the percentage by mass of aluminum and magnesium in the sample, we need to perform a few stoichiometric calculations based on the reaction of the alloy with hydrochloric acid and the production of hydrogen gas. The typical reactions for aluminum and magnesium with hydrochloric acid are:

Using the molar mass of hydrogen (1.008 g/mol), we can calculate the moles of hydrogen gas produced using the equation:

moles H₂ = 0.998 g / (2 * 1.008 g/mol) = 0.495 moles

The mole ratio of Al to H₂ in the reaction is 2:3, and for Mg to H₂ is 1:1. We can calculate the mass of aluminum or magnesium that would produce 0.495 moles of hydrogen.

For aluminum:

moles Al = (2/3) * moles H₂ = (2/3) * 0.495 = 0.330 moles

For magnesium:

moles Mg = moles H₂ = 0.495 moles

Now, we calculate the mass:

mass Al = moles Al * atomic mass Al = 0.330 moles * 26.98 g/mol = 8.904 g

mass Mg = moles Mg * atomic mass Mg = 0.495 moles * 24.31 g/mol = 12.033 g

These numbers are hypothetical maximums if the sample was 100% Al or Mg, and do not add up to 9.87 g. Therefore, we need to set up a system of equations considering the total mass of the alloy and the mass of hydrogen produced to find the exact mass of Al and Mg in the sample. This requires more information or a different approach that accounts for the actual reaction stoichiometry with the alloy sample, which isn't provided here.

The alloy consists of: 62.2% aluminium, 37.8% magnesium

To calculate the percentage by mass of each metal (aluminium and magnesium) in the alloy, we need to use the stoichiometry of their reactions with hydrochloric acid and the information given about the mass of the hydrogen gas produced.

Step 1: Write the balanced chemical equations for the reactions

For aluminium:

[tex]\[ 2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2 \][/tex]

For magnesium:

[tex]\[ \text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \][/tex]

Step 2: Determine the moles of hydrogen gas produced

The molar mass of hydrogen gas [tex](\(\text{H}_2\))[/tex] is:

[tex]\[ \text{Molar mass of H}_2 = 2 \times 1.008 = 2.016 \, \text{g/mol} \][/tex]

The number of moles of hydrogen gas produced is:

[tex]\[ \text{Moles of H}_2 = \frac{0.998 \, \text{g}}{2.016 \, \text{g/mol}} \]\[ \text{Moles of H}_2 = 0.495 \, \text{mol} \][/tex]

Step 3: Relate moles of hydrogen gas to moles of metals

Let  x be the mass of aluminium and  y be the mass of magnesium in the alloy.

From the reactions:

- Aluminium produces 3 moles of [tex]\(\text{H}_2\)[/tex] per 2 moles of [tex]\(\text{Al}\)[/tex]:

[tex]\[ 2\text{Al} \rightarrow 3\text{H}_2 \]\[ \frac{3}{2} \text{moles of H}_2 \text{ per mole of Al} \]\[ 1 \text{ mole of Al} \rightarrow \frac{3}{2} \text{ moles of H}_2 \][/tex]

- Magnesium produces 1 mole of [tex]\(\text{H}_2\) per mole of \(\text{Mg}\):[/tex]

[tex]\[ \text{Mg} \rightarrow \text{H}_2 \]\[ 1 \text{ mole of Mg} \rightarrow 1 \text{ mole of H}_2 \][/tex]

Thus, the moles of hydrogen gas produced by aluminium and magnesium are:

[tex]\[ \text{Moles of H}_2 \text{ from Al} = \frac{3}{2} \times \frac{x}{26.98} \]\[ \text{Moles of H}_2 \text{ from Mg} = \frac{y}{24.305} \][/tex]

The total moles of hydrogen gas produced:

[tex]\[ \frac{3}{2} \times \frac{x}{26.98} + \frac{y}{24.305} = 0.495 \][/tex]

Step 4: Set up the mass equation for the alloy

The total mass of the alloy is:

[tex]\[ x + y = 9.87 \, \text{g} \][/tex]

Step 5: Solve the system of equations

We have two equations:

[tex]1. \[ \frac{3}{2} \times \frac{x}{26.98} + \frac{y}{24.305} = 0.495 \]\\[/tex]

2.  x + y = 9.87

Let's solve these equations step by step.

First, express  y  in terms of x  from the second equation:

[tex]\[ y = 9.87 - x \][/tex]

Substitute this into the first equation:

[tex]\[ \frac{3}{2} \times \frac{x}{26.98} + \frac{9.87 - x}{24.305} = 0.495 \][/tex]

Simplify and solve for x :

[tex]\[ \frac{3x}{2 \times 26.98} + \frac{9.87 - x}{24.305} = 0.495 \]\[ \frac{3x}{53.96} + \frac{9.87 - x}{24.305} = 0.495 \]\[ \frac{3x}{53.96} + \frac{9.87}{24.305} - \frac{x}{24.305} = 0.495 \]\[ \frac{3x}{53.96} - \frac{x}{24.305} = 0.495 - \frac{9.87}{24.305} \][/tex]

Calculate the constant term:

[tex]\[ 0.495 - \frac{9.87}{24.305} = 0.495 - 0.406 = 0.089 \][/tex]

Combine the \( x \)-terms:

[tex]\[ \frac{3x}{53.96} - \frac{x}{24.305} = 0.089 \]\[ x \left( \frac{3}{53.96} - \frac{1}{24.305} \right) = 0.089 \][/tex]

Calculate the coefficient of \( x \):

[tex]\[ \frac{3}{53.96} = 0.0556 \]\[ \frac{1}{24.305} = 0.0411 \]\[ 0.0556 - 0.0411 = 0.0145 \][/tex]

Now, solve for x :

[tex]\[ x \times 0.0145 = 0.089 \]\[ x = \frac{0.089}{0.0145} \]\[ x = 6.14 \, \text{g} \][/tex]

Now, find y :

[tex]\[ y = 9.87 - x \]\[ y = 9.87 - 6.14 \]\[ y = 3.73 \, \text{g} \][/tex]

Step 6: Calculate the percentage by mass of each metal

[tex]\[ \% \, \text{Al} = \frac{6.14 \, \text{g}}{9.87 \, \text{g}} \times 100 = 62.2\% \]\[ \% \, \text{Mg} = \frac{3.73 \, \text{g}}{9.87 \, \text{g}} \times 100 = 37.8\% \][/tex]

Answer:

D (or E If properly listed to include the active site option)

Explanation:

A. Is true

Enzymes are organically biochemical catalyst and thus they can speed up the rate of chemical reaction in the body

B is true

They are catalysts as said earlier

C is true

They have active sites. An enzyme does not act on all substrates. They have particular group on which they can act. For example, we have carbohydrates enzymes that act on carbohydrates substrate only. This enzymes have no business acting on a protein substrate.

D. Enzymes are proteins

One of the important characteristics of enzymes is that they are protenious in nature

E. This is wrong. Enzymes like any over catalyst are not consumed in the course of the biochemical reaction

Enzymes are biological catalysts that speed up chemical reactions. These are proteins and have an 'active site' where reactions occur. However, they are not consumed during the reaction.

In context to your question about enzymes, they serve specific roles as biological accelerators or catalysts, significantly boosting the rate of a chemical reaction. This property is showcased under option A. They are indeed classified as proteins (option C), and they do have parts called active sites, where the substrate (the molecule upon which the enzyme acts) binds (option B).

However, option D posits that enzymes are consumed during the reaction, which is incorrect. Unlike many catalysts in non-biological reactions that get consumed during the reaction, enzymes remain unaffected by the reaction. They don't exhaust or alter in a reaction and are available to facilitate other reactions once the process is finished.

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Phosphoric acid is a triprotic acid that can donate three protons in solution, forming three anions. The pH of a 0.100 M solution is approximately 1.0, assuming it only ionizes once. The concentrations of the formed species are estimated to be highest for H2PO4⁻ and much lower for HPO4²⁻ and PO4³⁻.

Phosphoric acid, H3PO4 is a triprotic acid, meaning it can donate three hydrogen ions in a solution. This results in the formation of three different species:  H2PO4⁻, HPO4²⁻, and PO4³⁻.

The estimated pH of a 0.100 M phosphoric acid solution will depend on the degree of dissociation, but for the first ionization, we can approximate it using the expression pH=-log[H⁺], where [H⁺] is the hydronium ion concentration. Given that a 0.100 M phosphoric acid solution ionizes mostly once, we have [H⁺]≈0.100 M, leading to an estimated pH around 1.0.

Next, the concentrations of the species in equilibrium can be calculated without exact Kb or Ka values as long as we make the approximation that dissociation after the first hydrogen ion is minimal in a dilute solution like 0.100 M. In this case, we will assume [H2PO4⁻]≈0.100 M and [HPO4²⁻] and [PO4³⁻] will be much less than [H2PO4⁻].

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Answer:

The final temperature of the mixture is 21.4°C

Explanation:

Specific heat capacity of water = 1 cal/g°c

Heat loss by water  = 77.1 g X 1 cal/g°c X 50.9°c = 3924.39 Cal

Latent heat of fusion of ice = 79.7 g⁻¹

Heat required to melt ice at 0°c= 22.5 g X 79.7 g⁻¹ = 1793.25 Cal

Heat gained by ice from water at a higher temperature, T°c = 22.5 X 1 X T  

= 22.5T

Also heat lost by water = 77.1  X 1  X (50.9-T)cal

By calorimetric principle

Heat lost by a hot body = heat gained by a cold body

77.1  X 1  X (50.9-T) = 22.5T + 1793.25

3924.39 -77.1T = 22.5T + 1793.25

99.6T = 3924.39 - 1793.25

T = 2131.14/99.6

T = 21.4°C

Therefore, the final temperature of the mixture is 21.4°C

Answer: (a) NH3 + H20 ⇔ NH4⁺ + OH⁻ Kb =  [tex]\frac{ [NH4][OH]}{[NH3]}[/tex]

(b) CN⁻ + H20 ⇔ HCN⁺ + OH⁻ Kb =  [tex]\frac{ [HCN][OH]}{[CN]}[/tex]

(c) C5H5N + H20 ⇔ C5H5NH⁺ + OH⁻ Kb = [tex]\frac{[C5H5NH][OH]}{ [C5H5N]}[/tex]

(d) C6H5NH2 + H20 ⇔ C6H5NH3⁺ + OH⁻ Kb = [tex]\frac{[C6H5NH3][OH]}{[C6H5NH2] }[/tex]

Explanation: There are ways to calculate the strength of acids and bases. The pH is more commom but there is pKa, pKb, Ka and Kb. Ka and Kb are the dissociation constant for acids and bases, respectively.

Using the balanced equation of dissociation, Kb is calculated by dividing the concentration of the products with the concentration of the reagent.

The reactions and Kb equilibrium expressions for NH₃, CN⁻, pyridine (C₅H₅N), and aniline (C₆H₅NH₂) as bases in water are summarized including all necessary chemical equations and equilibrium expressions.

The following reactions and corresponding Kb equilibrium expressions for NH₃, CN⁻, pyridine (C₅H₅N), and aniline (C₆H₅NH₂) acting as bases in water are as follows:

Answer:

Synergism

Explanation:

This is an example of Synergism. Synergism is nothing but working out of two medicines together.

Examples of medical synergies are when doctors treat microbial heart infections with ampicillin and Gentamicin and when people with cancer undergo radiation and chemotherapy or more than one chemotherapy drug at a time.

Answer:

To regulate the gas pressure in the lighter tank and avoid build-up of pressure.  

Explanation:

We soak the lighter in the water bath to achieve uniform heating and reduce the fire hazard inherent in direct heating methods, ensuring a safer laboratory environment.

We soak the lighter in the water bath to ensure uniform heating of the reaction mixture with less fire hazard. Direct heating on a Bunsen burner or hot plate can cause uneven temperatures and increased risks of fire or overheating. A water bath is especially beneficial when a chemical reaction must be heated for a certain time to occur. It provides a stable and consistent heat source, which is safer and can prevent accidents in the lab. By using a water bath, we also prevent the lighter from getting excessively hot, which could cause it to malfunction or pose a safety risk.

Answer:

molecules

Explanation:

Answer:

5.6 × 10⁻¹⁷ J

Explanation:

We know that the work done by an electric field, E on an electric charge, q moving a distance, d is W = qEdcosθ. From above, q = 3.2 × 10⁻¹⁹ C, d = 20 cm = 2 × 10⁻² m, E = V/d = 175V /2 × 10⁻² m = 8750 V/m and θ = 0 since the α particle moves in the same direction as the electric field. So W = qEdcosθ = 3.2 × 10⁻¹⁹ C × 8750 V/m × 2 × 10⁻² m × cos0  = 5.6 × 10⁻¹⁷ J. We know that the work done by the electric field on the charge W = ΔK the change in kinetic energy of the charge. So, W = 1/2m(v₂² - v₁²) where v₁ = initial velocity and v₂ = final velocity. Since the charge is at rest at the positive plate, v₁ = 0. So, W = 1/2mv₂² = K which is the kinetic energy of the particle after moving the distance of 20 cm between the plates. So K = W = 5.6 × 10⁻¹⁷ J

Final answer:

The final temperature of the mixture will be 15.44°C.

Explanation:

In this problem, we can use the principle of conservation of energy to find the final temperature of the mixture. The heat gained by the ice cubes is equal to the heat lost by the water.

To find the heat gained by the ice cubes, we can use the formula:

Q = m × c × ΔT

where Q is the heat gained, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Since the ice cubes are at 0°C, the heat gained is:

Q = 4 × (53.5 g) × (4.184 J/(g°C)) × (0°C - T-f)

where T-f is the final temperature.

To find the heat lost by the water, we can use the formula:

Q = m × c × ΔT

where Q is the heat lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Since the water is at 75°C, the heat lost is:

Q = (115 g) × (4.184 J/(g°C)) × (T-f - 75°C)

Since no heat is lost to the surroundings, the heat gained by the ice cubes is equal to the heat lost by the water. Setting the two equations equal to each other, we can solve for T-f:

4 × (53.5 g) × (4.184 J/(g°C)) × (0°C - T-f) = (115 g) × (4.184 J/(g°C)) × (T-f - 75°C)

Simplifying the equation gives:

14.0816 × T-f = 496.82 - 18.0714 × T-f

32.1526 × T-f = 496.82

T-f = 15.44°C

Answer:

we only see parts of the lit side as the moon goes around the earth

Explanation:

Unlike the sun, the moon orbits the Earth. This is the reason why we see the different phases of the moon. The reflection of the moon is being illuminated back to us with the help of the sun. So, as the moon circles the Earth, we only see parts of the lit side. Such changes helps us see the moon in different phases such as the Third Quarter, Crescent, New Moon, Full Moon, etc.

For example, during "Full Moon," the moon's entire face is lit up by the sun. Thus, we see the entire moon's lit portion.

Thus, this explains the answer.

Answer:

The correct answer is "False".

Explanation:

The Elizabethan Era was a period of England's history that corresponds to  the reign of Queen Elizabeth I (1558–1603). The Elizabethan Era is considered a golden age of England's history, and part of the Renaissance Era (1300-1600). The Elizabethan view of life changed drastically from  the characteristic Medieval view of life. During Medieval times life was seen with a religiously perspective only, while during The Elizabethan Era more people start to view life with a scientific perspective.

THE ANSWER IS FALSE!

Answer: 0.147

Explanation:

[tex]K_p[/tex] is the constant of a certain reaction at equilibrium.

For the given chemical reaction:

     [tex]NH_4HS(s)\rightleftharpoons NH_3(g)+H_2S(g)[/tex]

at t= 0       0                                           0               0

at eqm                                                   p                p

Total pressure = p+p = 0.766 atm

2p= 0.766 atm

p= 0.383 atm

The expression of [tex]K_p[/tex] for above equation follows:

[tex]K_p={p_{NH_3}\times p_{H_2S}}[/tex]

We are given:

[tex]K_p={0.383\times 0.383[/tex]

[tex]K_p=0.147[/tex]

Thus [tex]K_{eq}[/tex] at this temperature is 0.147

Answer:

The first question is water vapor.

Explanation:

Water vapor is not a green house gas. Clouds are made of water vapor.

Answer:

Explanation:

The same charges repel each other while opposite charges attract each other. During electron-electron interaction repulsion take palace because the electron has negative charges. Nucleus has positive charges so the interaction between two nucleus results in the form of repulsion. When interaction takes place between nucleus and electron then attraction takes place between nucleus and electrons due to opposite charges.

          The formation of a bond that takes place due to the sharing of the electrons is known as a covalent bond and thus the covalent molecule is formed.

The forces between two atoms is the result of electron-electron repulsion, nucleus-nucleus repulsion, and nucleus-electron attraction. At the point of equilibrium, the attractive forces balance the repulsive forces. The most stable configuration of atoms exists at the point of minimum potential energy, when the atoms bond covalently and a covalent bond forms.

The force of attraction between atoms leads to the formation of covalent bonds. The force of attraction is defined by the magnitude of oppositely charged ions bonded with one another. There are various forces of attraction in between molecules which are studied in chemistry. These are:

Answer:

The answer to this is the concentration of glucose (C6H12O6) in mg/dL = 95.48 mg/dL

Explanation:

To solve this we list out the known variables thus

Measured concentration of  glucose  (C6H12O6) = 5.3mM per liter

The molar mass of glucose  = 180.156 g/mol

From the above, it is seen that one mole of glucose contains 180.156 grams of  C6H12O6 therefore 5.3 mM which is 5.3 × 10⁻³ moles contains

5.3 × 10⁻³ moles × 180.156 g/mol = 0.9548 grams of glucose

Also 1 d L = 0.1 L  or 1 L = 10 dL and 1 mg = 1000 g, hence

thus 0.9548 grams per liter is equivalent to 1000/10 × 0.9548 milligrams per dL or 95.48 mg/dL

Answer:

The answer to this question is 45.33 moles of aluminum sulfate is produced when 125 moles of aluminum hydroxide and 136 moles of sulfuric acid react

Explanation:

To solve this, we write out the chemical equation of he reaction thus

Al(OH)3(s) + 3 H2SO4(aq) -----> Al2 (SO4)3(aq) + 6 H2O(l)

here it is seen that one moles of aluminum hydroxide reacts with three moles of sulfuric to produce one mole of aluminum sulfate and six moles of water

hence

136 moles of sulfuric acid reacts with 136/3 or 45.33 moles of aluminum hydroxide to produce 136/3 or 45.33 moles of aluminum sulfate and 2× 136 moles of water

Hence the amount in moles of aluminum sulfate produced is 45.33 moles

Final answer:

Using the balanced chemical equation 3 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O, and knowing that aluminum hydroxide is the limiting reactant, 125 moles of aluminum hydroxide will produce 41.67 moles of aluminum sulfate.

Explanation:

The question asks how many moles of aluminum sulfate will be produced when reacting 125 moles of aluminum hydroxide with 136 moles of sulfuric acid. To answer this, we need the balanced chemical equation:

3 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O

The stoichiometry of the reaction shows that 3 moles of aluminum hydroxide react with 3 moles of sulfuric acid to produce 1 mole of aluminum sulfate. Since there are more moles of sulfuric acid present, aluminum hydroxide is the limiting reactant. Therefore, we can calculate the moles of aluminum sulfate produced by dividing the moles of aluminum hydroxide by 3, which gives us:
125 moles Al(OH)3 ÷ 3 = 41.67 moles Al2(SO4)3 (rounded to two decimal places as per the significant figures in the provided moles of reactants).

To calculate the vapor pressure of the solution, we need to consider Raoult's law. According to Raoult's law, the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent. In this case, the solvent is water and the solute is K3PO4. To find the mole fraction of water, we divide the moles of water by the total moles of solute and solvent. Using the given values, the vapor pressure of the solution at 55 °C is 101.0 torr.

To calculate the vapor pressure of the solution, we need to consider Raoult's law. According to Raoult's law, the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent. In this case, the solvent is water and the solute is K3PO4. To find the mole fraction of water, we divide the moles of water by the total moles of solute and solvent. Using the given values, we have:

Moles of water = 0.941 mol

Total moles of solute and solvent = 0.159 mol K3PO4 + 0.941 mol H2O = 1.1 mol

Mole fraction of water = (0.941 mol) / (1.1 mol) = 0.855

Now, we can use the mole fraction of water and the vapor pressure of pure water to calculate the vapor pressure of the solution:

Vapor pressure of the solution = mole fraction of water * vapor pressure of pure water

= 0.855 * 118.1 torr

= 101.0 torr

Therefore, the vapor pressure of the solution at 55 °C is 101.0 torr.

Using Raoult's Law, the vapor pressure of the solution at 55°C is approximately 100.93 torr.

To calculate the vapor pressure of the solution, we will use Raoult's Law, which states:

[tex]\[ P_{\text{solution}} = X_{\text{solvent}} \cdot P^{\star}_{\text{solvent}} \][/tex]

where:

- [tex]\( P_{\text{solution}} \)[/tex] is the vapor pressure of the solution,

- [tex]\( X_{\text{solvent}} \)[/tex] is the mole fraction of the solvent,

- [tex]\( P^{\star}_{\text{solvent}} \)[/tex] is the vapor pressure of the pure solvent.

Given data:

- Mole fraction of water [tex](\( X_{\text{H2O}} \))[/tex] in the solution:

 [tex]\[ X_{\text{H2O}} = \frac{n_{\text{H2O}}}{n_{\text{H2O}} + n_{\text{K3PO4}}}[/tex]

[tex]= \frac{0.941}{0.159 + 0.941} \\\\= \frac{0.941}{1.1} \\ \\ \approx 0.855[/tex]

- Vapor pressure of pure water at 55 °C:

[tex]\[ P^{\star}_{\text{H2O}} = 118.1 \text{ torr} \][/tex]

Now, calculate the vapor pressure of the solution:

[tex]\[ P_{\text{solution}} = X_{\text{H2O}} \cdot P^{\star}_{\text{H2O}} \][/tex]

[tex]\[ P_{\text{solution}} = 0.855 \times 118.1 \text{ torr} \][/tex]

[tex]\[ P_{\text{solution}} \approx 100.93 \text{ torr} \][/tex]

Therefore, the vapor pressure of the solution at 55 °C is approximately 100.93 torr.

Answer:

The answer to your question is below

Explanation:

The specific heat is a physical property equal to the amount of heat necessary to increase the temperature of 1 gram of a substance by one degree celsius.

The lower the specific heat, the lower the amount of heat to increase the temperature 1°C, the higher the specific heat, the higher the amount of heat necessary to increase the temperature by 1°C.

The specific heat of copper is 0.093 cal/g°C

The specific heat of water is 1 cal/g°C.

That is why is necessary more heat to warm water.

More energy to increase the temperature of 100 grams of liquid water by one degree Celsius than it does 100 grams of copper metal due to higher specific heat.

Specific heat refers to the amount of heat needed to raise the temperature of 1 gram of a substance by 1 degree Celsius. Water has a high specific heat which means it takes more energy to increase the temperature of water compared to other substances like metals.

So we can conclude that more energy is needed to increase the temperature of 100 grams of liquid water by one degree Celsius than copper metal because of higher specific heat of water.

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