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jok3333 [9.3K]
2 years ago
5

How many grams of sodium chloride are present in a 0.75 M solution with a volume of 500.0 milliliters?

Chemistry
1 answer:
LUCKY_DIMON [66]2 years ago
6 0
The units for molarity is moles of solute per liter of solution which means if you multiply the molarity of a solution by its volume you get how many moles of solute are in the solution. (0.75Mx0.5L=0.375mol NaCl) Then you can multiply the moles of sodium chloride (0.375 mol) by its molar mass (58.45 g/mol) to get 21.92g of sodium chloride. That means there is 21.92 grams of sodium chloride in 500mL of 0.75M solution. I hope this helps. Let me know if anything is unclear.
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<u>Answer:</u> The standard heat for the given reaction is -138.82 kJ

<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^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

For the given chemical reaction:

4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)

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

\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]

We are given:

\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ

Hence, the standard heat for the given reaction is -138.82 kJ

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Answer:

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Explanation:

The reaction is:

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Let's determine the amount of acid.

M are the moles contained in 1 L of solution or it can be mmoles that are contained in 1 mL of solution

M = mmol /mL

M . mL = mmol

We replace: 8.3 mL . 9.9 M = 82.17 mmoles

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4 moles of sulfuric acid can make 2 moles of alum

By the way, 0.082 moles of acid may produce ( 0.082 . 2) /4 = 0.041085 moles.

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