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Burka [1]
4 years ago
12

A 6.0 Liter solution is made with 2 moles of solute. What is the molarity of the solution?

Chemistry
1 answer:
kodGreya [7K]4 years ago
3 0
The formula for molarity is: mol/L. so it would be 2.0 mol/ 6.0 L.

the answer is: .33 M

hope this helps!
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When an object reflects all light, you will see white light.

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What state of matter having a definite volume but no definite shape
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The state of matter is liquid.
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What is the ph of a solution that is 0.100 m hio3and 0.100 m naio3?
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4 years ago
When NH3(g) reacts with N2O(g) to form N2(g) and H2O(g), 105 kcal of energy are evolved for each mole of NH3(g) that reacts. Wri
Ganezh [65]

Answer : The balanced chemical equation is,

2NH_3(g)+3N_2O(g)\rightarrow 4N_2(g)+3H_2O(g)+210kcal

Explanation :

Balanced chemical equation : It is defined as the number of atoms of individual elements present on the reactant side must be equal to the number of atoms of individual elements present on product side.

The given unbalanced chemical reaction is,

NH_3(g)+N_2O(g)\rightarrow N_2(g)+H_2O(g)+105kcal

This chemical reaction is an unbalanced reaction because in this reaction, the number of atoms of individual elements are not balanced.

In order to balanced the chemical reaction, the coefficient 2 is put before the NH_3, the coefficient 3 is put before the N_2O\text{ and }H_2O and the coefficient 4 is put before the N_2.

The energy evolved in this reaction = 105Kcal\times 2=210Kcal

Thus, the balanced chemical reaction will be,

2NH_3(g)+3N_2O(g)\rightarrow 4N_2(g)+3H_2O(g)+210kcal

7 0
3 years ago
An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?
umka2103 [35]

<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

K_b = molal boiling point elevation constant = 0.51°C/m

m_{solute} = Given mass of solute (CsCl) = 8.00 g

M_{solute} = Molar mass of solute (CsCl) = 168.4  g/mol

W_{solvent} = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC

Hence, the boiling point of solution is 100.53

6 0
3 years ago
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