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sveticcg [70]
3 years ago
7

Using the following thermochemical data, what is the change in enthalpy for the following reaction: 3H2(g) + 2C(s) + ½O2(g) C2H5

OH(l) , ΔH = –1367 kJ/mol , ½ ,
Chemistry
1 answer:
Tamiku [17]3 years ago
6 0
Given: 

<span>3H2(g) + 2C(s) + ½O2(g)     </span>→   <span>C2H5OH(l) , ΔH = –1367 kJ/mol 

Change is enthalpy is mathematically expressed as enthalpy of product minus enthalpy of reactant. If value of enthalpy change is negative, it suggest reaction is exothermic. If enthalpy change is positive, it suggest reaction is endothermic.

In present case, enthalpy change is -1367 kJ/mol.</span>
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How many grams of nitric acid HNO₃, are required to neutralize (completely react with) 4.30 grams of Ca(OH)2 according to the ac
Brrunno [24]

Answer:

7.32g of HNO3 are required.

Explanation:

1st) From the balanced reaction we know that 2 moles of HNO3 react with 1 mole of Ca(OH)2 to produce 2 moles of H2O and 1 mole of Ca(NO3)2.

From this, we find that the relation between HNO3 and Ca(OH)2 is that 2 moles of HNO3 react with 1 mole of Ca(OH)2.

2nd) This is the order of the relations that we have to use in the equation to calculate the grams of nitric acid:

• starting with the 4.30 grams of Ca(OH)2.

,

• using the molar mass of Ca(OH)2 (74g/mol).

,

• relation of the 2 moles of HNO3 that react with 1 mole of Ca(OH)2 .

,

• using the molar mass of HNO3 (63.02g/mol).

4.30g\text{ Ca\lparen OH\rparen}_2*\frac{1\text{ mol Ca\lparen OH\rparen}_2}{74g\text{ Ca\lparen OH\rparen}_2}*\frac{2\text{ moles HNO}_3}{1\text{ mole Ca\lparen OH\rparen}_2}*\frac{63.02g\text{ HNO}_3}{1\text{ mole HNO}_3}=7.32g\text{ HNO}_3

So, 7.32g of HNO3 are required.

4 0
1 year ago
You need 270 ml of a 65% alcohol solution. on hand, you have a 90% alcohol mixture. how much of the 90% alcohol mixture and pure
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You must add 75 mL water to 195 mL 90 % alcohol to make 270 mL of 65 % alcohol.

<em>Step 1.</em> Calculate the volume of 90 % alcohol needed

You can use the dilution formula

<em>V</em>1×<em>C</em>1 = <em>V</em>2×<em>C</em>2

where

<em>V</em>1 and<em> V</em>2 are the volumes of the two solutions

<em>C</em>1 and <em>C</em>2 are the concentrations

You can solve the above formula to get

<em>V</em>2 = <em>V</em>1 × <em>C</em>1/<em>C</em>2

<em>V</em>1 = 270 mL; <em>C</em>1 = 65 %

V2 = ?; _____<em>C</em>2 = 90 %

∴<em>V</em>2 = 270 mL × (65 %/90 %) = 195 mL

You need 195 mL of 90 % alcohol to make 270 mL of 65 % RA

<em>Step 2</em>. Calculate the amount of water to add.

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