THE ASTHENOSPHERE IS CONSIDERED TO BE A SOFT SOLID.True or False

Suppose that 2.14 grams of H2SO4 is mixed with enough water to make 225 mL of solution. Determine the Molarity (M) of the soluti

Alik [6]

Answer:

0.0970 M

Explanation:

Remember this equation:

mol/M x V

Convert it so that you can get M.

M=mol/V

Convert the 2.14 grams of H2SO4 into mols

=0.0218

Convert mL to L

225/1000

=0.225

Plug it in.

0.0218/0.225

=0.0970 M

6 0

2 years ago

Which part of the circulatory system contains blood rich in carbon dioxide

horrorfan [7]

The answer is the veins

8 0

3 years ago

When 2.16g of H2 reacts with excess O2 by the following equation, 258 kJ of heat are released. What is the change of enthalpy as

alekssr [168]

Answer:

-241 kJ/mol

Explanation:

Let's consider the reaction of hydrogen with excess oxygen to form water.

2 H₂ + O₂ ⟶ 2 H₂O

When 2.16g of hydrogen reacts with excess oxygen, 258 kJ of heat are released, that is, Q = -258 kJ. Considering that the molar mass of hydrogen is 2.02 g/mol, the change of enthalpy associated with the reaction of 1.00 mol of hydrogen gas is:

ΔH° = -258 kJ/2.16 g × (2.02 g/1.00 mol) = -241 kJ/mol

8 0

2 years ago

Convert 1.8 moles of potassium to atoms

castortr0y [4]

703.7693999989732 i believe if i did my math right

3 0

3 years ago

Commercial hydrochloric acid (HCl) is typically labeled as being 38.0 % (weight %). The density of HCl is 1.19 g/mL. a) What is

Strike441 [17]

Answer:

For a: The molarity of commercial HCl solution is 12.39 M.

For b: The molality of commercial HCl solution is 16.79 m.

For c: The volume of commercial HCl solution needed is 2.42 L.

Explanation:

We are given:

Mass % of commercial HCl solution = 38 %

This means that 38 grams of HCl is present in 100 grams of solution.

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

$Density=\frac{Mass}{Volume}$

Density of HCl solution = 1.19 g/mL

Mass of solution = 100 g

Putting values in above equation:

$1.19g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=84.034mL$

For a:

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

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = ?

Molar mass of HCl = 36.5 g/mol

Volume of solution = 84.034 mL

Mass of HCl = 38 g

Putting values in above equation, we get:

$\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 84.034}\\\\\text{Molality of commercial HCl solution}=12.39M$

Hence, the molarity of commercial HCl solution is 12.39 M.

For b:

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

$Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (HCl) = 38 g

$M_{solute}$ = Molar mass of solute (HCl) = 36.5 g/mol

$W_{solvent}$ = Mass of solvent = 100 - 38 = 62 g

Putting values in above equation, we get:

$\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 62}\\\\\text{Molality of commercial HCl solution}=16.79m$

Hence, the molality of commercial HCl solution is 16.79 m.

For c:

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

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted solution

We are given:

$M_1=6M\\V_1=5.00L\\M_2=12.39M\\V_2=?L$

Putting values in above equation, we get:

$6\times 5=12.39\times V_2\\\\V_2=2.42L$

Hence, the volume of commercial HCl solution needed is 2.42 L.

5 0

2 years ago