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Delicious77 [7]
2 years ago
8

Given: C + O2 → CO2 Bond Bond Energy (kJ/mol) C=O 799 O=O 494 Calculate the enthalpy change for the chemical reaction. The chang

e in enthalpy for the given reaction is kilojoules.
Chemistry
1 answer:
aliya0001 [1]2 years ago
3 0

Answer:

-1104 kJ/mol

Explanation:

The change in the enthalpy of a reaction is equal to the difference between: the sum of the enthalpy changes of the bonds broken and the sum of the enthalpy changes of the bonds formed.

The bonds broken correspond to the cleavage of bonds of the reactants, the bonds formed correspond to the bonds of the products:

  • we only break oxygen O=O bond, since carbon is not bonded to anything;
  • we form two C=O bonds in carbon dioxide.

Therefore, the enthalpy change is calculated by:

\Delta H^o = 1 mol\cdot 494 kJ/mol - 2 mol\cdot 799 kJ/mol = -1104 kJ/mol

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Calculate how many grams of the product form when 16.7 g of calcium metal completely reacts. Assume that there is more than enou
swat32

39.96 g product form when 16.7 g of calcium metal completely reacts.

<h3>What is the stoichiometric process?</h3>

Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Equation:

Ca(s) + Cl_2(g) → CaCl_2(s)

In this case, for the undergoing reaction, we can compute the grams of the formed calcium chloride by noticing the 1:1 molar ratio between calcium and it (stoichiometric coefficients) and using their molar mass of 40 g/mol and 111 g/mol by using the following stoichiometric process:

m_{ca_C_l_2}= 16.7 g Ca x \frac{1 mol \;of \;Ca}{40g Ca} x \frac{1 mol \;of \;CaCl_2}{1 mol \;Ca} x \frac{111g of \;CaCl_2}{1 mol \;CaCl_2}

m_{ca_C_l_2} = 39.96 g

Hence, 39.96 g product form when 16.7 g of calcium metal completely reacts.

Learn more about the stoichiometric process here:

brainly.com/question/15047541

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4 0
2 years ago
How much 6.0 m hno3 is needed to neutralize 39ml of 2 m koh
Sever21 [200]

Answer:

13mL

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

HNO3 + KOH —> KNO3 + H2O

From the balanced equation above, we obtained the following data:

Mole ratio of the acid (nA) = 1

Mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question.

This includes the following:

Molarity of the acid (Ma) = 6M

Volume of the acid (Va) =?

Volume of the base (Vb) = 39mL

Molarity of the base (Mb) = 2M

Step 3:

Determination of the volume of the acid.

Using the equation:

MaVa/MbVb = nA/nB, the volume of the acid can be obtained as follow:

MaVa/MbVb = nA/nB

6 x Va / 2 x 39 = 1/1

Cross multiply to express in linear form

6 x Va = 2 x 39

Divide both side by 6

Va = (2 x 39)/6

Va = 13mL

Therefore, the volume of the acid (HNO3) needed for the reaction is 13mL

5 0
3 years ago
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