At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. What partial pressure of He would give a solubility of 0.730 M

Answers

Answer 1

Answer:

We have that from the Question, it can be said that The partial pressure of He would give a solubility of 0.730 M is

P_2=4.7atm

From the Question we are told

At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. What partial pressure of He would give a solubility of 0.730 M

Generally the equation for constant temperature is mathematically given as

$(C_2)/(C_1)=(P_2)/(P_1)\n\nTherefore\n\nP_2=(P_1C_1)/(C_1)\n\nP_2=(0.22*1.7)/(0.080)\n\nP_2=4.7atm\n\n$

Therefore

The partial pressure of He would give a solubility of 0.730 M is

P_2=4.7atm

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Answer 2

Answer:

Answer: Partial pressure of He that would give a solubility of 0.730 M is 15.5 atm

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:

$C_(He)=K_H* p_(liquid)$

where,

$K_H$ = Henry's constant =?

$p_(He)$ = partial pressure = 1.7 atm

Putting values in above equation, we get:

$0.080=K_H* 1.7atm\n\nK_H=0.047Matm^(-1)$

To find partial pressure of He would give a solubility of 0.730 M

$0.730=0.047Matm^(-1)* p_(liquid)$

$p_(liquid)=15.5atm$

Thus partial pressure of He that would give a solubility of 0.730 M is 15.5 atm

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Be sure to answer all parts.Calculate the percent composition by mass (to 4 significant figures) of all the elements in calcium
phosphate (Ca3(PO4)2), a major component of bone.
% Ca
%P
% 0

Answers

Answer:

38.7%

41.3%

20%

Explanation:

The percentage composition helps to know the what percent of the total mass of a compound is made up of each of the constituent elements or groups.

To solve this problem:

find the formula mass by adding the atomic masses of the atoms that makes up the compound.
place the mass contribution of the element or group to the formula mas and multiply by 100;

Compound:

Ca₃(PO₄)₂

Formula mass = 3(40) + 2[31 + 4(16)]

= 120 + 2(95)

= 120 + 190

= 310

%C = $(3(40))/(310)$ x 100 = 38.7%

%P = $(8(16))/(310)$ x 100 = 41.3%

%O = $(2(31))/(310)$ x 200 = 20%

Be sure to answer all parts. Industrially, hydrogen gas can be prepared by combining propane gas (c3h8) with steam at about 400°c. The products are carbon monoxide (co) and hydrogen gas (h2). (a) write a balanced equation for the reaction. Include phase abbreviations. (b) how many kilograms of h2 can be obtained from 8.31 × 103 kg of propane

Answers

Balanced chemical reaction:

C₃H₈(g) + 3H₂O(g) → 3CO(g) + 7H₂(g).

M(C₃H₈) = 44.1 g/mol; molar mass of propane.

M(H₂) = 2 g/mol; molar mass of hydrogen.

From balanced chemical reaction: n(C₃H₈) : n(H₂) = 1 : 7.

7m(C₃H₈) : M(C₃H₈) = m(H₂) : M(H₂).

7·8310 kg : 44.1 g/mol = m(H₂) : 2 g/mol.

m(H₂) = 2638.09 kg; mass of hydrogen.

Answer: a) $C_3H_8(g)+3H_2O(g)\rightarrow 3CO(g)+7H_2(g)$

b) $2.64* 10^3kg$

Explanation:

a) According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$C_3H_8(g)+3H_2O(g)\rightarrow 3CO(g)+7H_2(g)$

b) $\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of propane}=(8.31* 10^6g)/(44.1g/mol)=0.188* 10^6moles$

$C_3H_8(g)+3H_2O(g)\rightarrow 3CO(g)+7H_2(g)$

According to stoichiometry:

1 mole of $C_3H_8$ gives 7 moles of $H_2$

Thus $0.188* 10^6moles$ moles of $C_3H_8$ will give = $(7)/(1)* 0.188* 10^6=1.32* 10^6moles$ of $H_2$

Mass of $H_2=moles* {\text {molar Mass}}=1.32* 10^6moles* 2g/mol=2.64* 10^6g=2.64* 10^3kg$

Thus $2.64* 10^3kg$ of $H_2$ can be obtained from $8.31* 10^3$ kg of propane

Balance the chemical equation given below, and determine the number of moles of iodine that reacts with 10.0 g of aluminum._____ Al(s) + _____ I2(s) → _____ Al2I6(s)

Answers

Answer:

1. 2Al + 3I2 —> Al2I6

2. 0.555mol of I2

Explanation:

1. Al + I2 —> Al2I6

Observing the above equation, there are 2 atoms of Al on the right side and 1 on the left side. To balance it, put 2 in front of Al as shown below:

2Al + I2 —>Al2I6

Also, there are 6 atoms of I on the right side and 2 on the left side. To balance it, put 3 in front I2 as shown below:

2Al + 3I2 —>Al2I6

2. Molar Mass of Al = 27g/mol

Mass of Al = 10g

n = Mass /Molar Mass

n = 10/27 = 0.37mol

From the equation,

2moles of Al reacted with 3 moles of I2.

Therefore, 0.37mol of Al will react with = (0.37 x 3)/2 = 0.555mol of I2

The balanced chemical equation is:

2 Al(s) + 3 I₂(s) → Al₂I₆(s)

0.557 moles of iodine react with 10.0 g of aluminum.

Let's consider the following unbalanced equation.

Al(s) + I₂(s) → Al₂I₆(s)

We will balance it using the trial and error method. We can get the balanced equation by multiplying Al by 2 (balance Al atoms) and I₂ by 3 (balance I atoms).

2 Al(s) + 3 I₂(s) → Al₂I₆(s)

The molar mass of Al is 26.98 g/mol. The moles corresponding to 10.0 g of Al are:

$10.0 g * (1mol)/(26.98g) = 0.371 mol$

The molar ratio of Al to I₂ is 2:3. The moles of I₂ that react with 0.371 moles of Al are:

$0.371 molAl * (3molI_2)/(2molAl) = 0.557 mol I_2$

The balanced chemical equation is:

2 Al(s) + 3 I₂(s) → Al₂I₆(s)

0.557 moles of iodine react with 10.0 g of aluminum.

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Which chemical equation follows the law of conservation of mass?

Answers

The chemical equation presented in option A follows the law of conservation of mass.

The principle of conservation of mass states, mass can neither be created nor destroyed but can be transformed from one form to another.

A reaction that follows the law of conservation of mass, must have equal number of moles each elements in reactants side and products side.

Only option A follows the law of conservation of mass;

$2LiOH \ + \ + H_2CO_3 \ ---> \ Li_2CO_3 \ + \ 2H_2O$

Thus, we can conclude that the chemical equation presented in option A follows the law of conservation of mass.

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Answer:

Option A

Explanation:

The expression that obeys the law of conservation of mass is choice A;

2LiOH + H₂CO₃ → Li₂CO₃ + 2H₂O

According to the law of conservation of mass; "in a chemical reaction, matter is neither created nor destroyed". By this law, mass is usually conserved.

The equation shows that mass is conserved because the number of moles of each specie is found on both sides

Number of moles

Li O H C

Reactants 2 5 4 1

Products 2 5 4 1

This shows that mass is indeed conserved.

Give an example of an element and an example of a compound.

Answers

Answer:

an element is a atom like titanium and a compound is like a water, glucose, alcohol and salt

Explanation:

the titanium is a element and water, glucose, alcohol, and salt those are a compound

How do you figure out the # of
neutrons?

Answers

Answer:

Subtracting the number of protons from the atomic mass.

Answer:

Subtract Protons

Explanation:

Since the vast majority of atom's mass is found it's protons and neutrons, Subtracting the Number of Protons ( i.e. the atomic number ) from the atomic mass will give you the calculated number of neutrons in a atom,

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