Ask question

Ask question

All categories

3 years ago

6

Ok! Water is added to 200. ml of a 2.0 M solution of CaCI2 to increase the volume of the solution to 450. ml. So what is the new

concentration?

1 answer:

dexar [7]3 years ago

7 0

Answer:

0.89M

Explanation:

Because moles of solute do not change, we can write

M1V1 = M2V2

M - molarity, V - volume of solution

2.0M*200mL = M2*450mL

M2 = 0.89 M

You might be interested in

At certain times during the process, the temperature increased. During these times, the heat that was absorbed took the form of

Ann [662]

Answer: kinetic, move faster, potential, change form

Explanation:

5 0

3 years ago

What property allows metals to be shaped into a musical instrument?

FinnZ [79.3K]

The correct answer is D) Malleability

8 0

3 years ago

Ameadow food chain is shown.

GalinKa [24]

The mouse is a Secondary consumer

7 0

3 years ago

Read 2 more answers

A _____ forecasts the weather using data collected from many sources.

Sav [38]

Meteorologist.............................

7 0

3 years ago

Given the following at 25C calculate delta Hf for HCN (g) at 25C. 2NH3 (g) +3O2 (g) + 2CH4 (g) ---> 2HCN (g) + 6H2O (g) delta

AysviL [449]

<u>Answer:</u> The $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]$

We are given:

$\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(NH_3)}=-80.3kJ/mol\\\Delta H_f_{(CH_4)}=-74.6kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ$

Putting values in above equation, we get:

$-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ$

Hence, the $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

8 0

3 years ago