What examples of a solid turning into a liquid

What mass of hcl gas must be added to 1.00 l of a buffer solution that contains [aceticacid]=2.0m and [acetate]=1.0m in order to

Paraphin [41]

PH of acidic buffer = pKa + log [CH₃COONa - HCl] / [CH₃COOH + HCl]
pKa of CH₃COOH = 4.74
Concentration of acetic acid in buffer = 2.0 M
Concentration of sodium acetate = 1.0 M
Concentration of HCl must add = x
pH = 4.74 + log (1-x) / (2+x) = 4.11
x = concentration of HCl must be added = 0.43 M
number of moles of HCl = M * V = 0.43 * 1 = 0.43 mol
mass of HCl must be added = 0.43 * 36.5 = 15.7 g

3 0

4 years ago

Iron (III) oxide and hydrogen react to form iron and water, like this: Fe 03(s)+3H9)2Fe(s)+3HO) At a certain temperature, a chem

belka [17]

The question is incomplete, here is the complete question:

Iron (III) oxide and hydrogen react to form iron and water, like this:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

At a certain temperature, a chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound Amount

Fe₂O₃ 3.95 g

H₂ 4.77 g

Fe 4.38 g

H₂O 2.00 g

Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

Answer: The value of equilibrium constant for given equation is $1.0\times 10^{-4}$

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

For hydrogen gas:

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

$\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M$

For water:

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

$\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M$

For the given chemical equation:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

The expression of equilibrium constant for above equation follows:

$K_{eq}=\frac{[H_2O]^3}{[H_2]^3}$

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

$K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}$

Hence, the value of equilibrium constant for given equation is $1.0\times 10^{-4}$

6 0

4 years ago

Please help! 30 points!

lisabon 2012 [21]

The person above me is correct

8 0

3 years ago

9. Let's say, you need to make 2.00 L of 0.05 M copper (II) nitrate solution. How many grams of

Vlad [161]

Answer:

18.76 g of copper II nitrate

Explanation:

Now recall that we must use the formula;

n= CV

Where;

n= number of moles of copper II nitrate solid

C= concentration of copper II nitrate solution

V= volume of copper II nitrate solution

Note that;

n= m/M

Where;

m= mass of solid copper II nitrate

M= molar mass of copper II nitrate

Thus;

m/M= CV

C= 0.05 M

V= 2.00 L

M= 187.56 g/mol

m= the unknown

Substituting values;

m/ 187.56 g/mol = 0.05 M × 2.00 L

m= 0.05 M × 2.00 L × 187.56 g/mol

m= 18.76 g of copper II nitrate

Therefore, 18.76 g of copper II nitrate is required to make 0.05 M solution of copper II nitrate in 2.00 L volume.

6 0

3 years ago

The concept of an ideal gas is used to explain(1) the mass of a gas sample(2) the behavior of a gas sample(3) why some gases are

Sauron [17]

Ideal gas is used to explain the behavior of a gas sample. In ideal gas law, the equation is PV=nRT where R has a constant of 0.0821L.atm/mol.K. It can explain the behaviior of a gas in three types: mole, mass and density. The ideal gas often observed in a high temperature with low pressure as to potential energy becomes less significant compared to the kinetic energy.

7 0

3 years ago