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Anastaziya [24]
2 years ago
10

How many moles of solute are present in .75 l of a .89 m (molar) solution?

Chemistry
1 answer:
never [62]2 years ago
8 0

0.6675 moles of solute are present in .75 l of a .89 m (molar) solution.

<h3>Define the molarity of a solution.</h3>

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution.

Given data:

V= 0.75

M=0.89

Molality = \frac{Moles \;solute}{Volume of solution in litre}

0.89 M=  \frac{Moles \;solute}{0.75 L}

Moles= 0.6675

Hence, 0.6675 moles of solute are present in .75 l of a .89 m (molar) solution.

Learn more about the moles here:

brainly.com/question/24081006

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An aluminum can holds 350 mL of gas at 0 C and 1.0 atm. what is the new volume if the can is heated to 10 C and the pressure ins
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Answer:

Final volume of the gas is 4.837mL

Explanation:

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To solve this question, we'll have to use combined gas equation which is the combination of all gas law I.e Charle's laws, Boyle's law, Pressure law etc.

According to combined gas equation,

(P1 × V1) / T1 = (P2 × V2) / T2

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Answer: The enthalpy of formation of SO_3 is  -396 kJ/mol

Explanation:

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The chemical equation for the combustion of propane follows:

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

\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(O_2(g))})]

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\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-297kJ/mol\\\Delta H^o_{rxn}=-198kJ

Putting values in above equation, we get:

-198=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta -297)+(1\times (0))]\\\\\Delta H^o_f_{(SO_3(g))}=-396kJ/mol

The enthalpy of formation of SO_3 is -396 kJ/mol

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