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3 years ago

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What result in a chemical change in matter

1 answer:

marishachu [46]3 years ago

3 0

Answer:

A chemical change results from a chemical reaction, while a physical change is when matter changes forms but not chemical identity. Examples of chemical changes are burning, cooking, rusting, and rotting. Examples of physical changes are boiling, melting, freezing, and shredding. Often, physical changes can be undone, if energy is input

Explanation:

Hope this helps :)

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CO(g) + 12 O2(g) → CO2(g)The combustion of carbon monoxide is represented by the equation above.(a) Determine the value of the s

devlian [24]

Answer : The standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The combustion of $CO$ will be,

$CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)$ $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $C(s)+\frac{1}{2}O_2(g)\rightarrow CO(g)$ $\Delta H_1=-110.5kJ/mol$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ/mol$

Now we are reversing reaction 1 and then adding both the equations, we get :

(1) $CO(g)\rightarrow C(s)+\frac{1}{2}O_2(g)$ $\Delta H_1=110.5kJ/mol$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ/mol$

The expression for enthalpy change for the reaction will be,

$\Delta H_{rxn}=\Delta H_1+\Delta H_2$

$\Delta H_{rxn}=(110.5)+(-393.5)$

$\Delta H_{rxn}=-283kJ/mol$

Therefore, the standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol

6 0

4 years ago

Describe how you would prepare 350 ml of 0.100 m c12h22o11 starting with 3.00l of 1.50 m c12h22o11

Leona [35]

To prepare 350 mL of 0.100 M solution from a 1.50 M solution, we simply have to use the formula:

M1 V1 = M2 V2

So from the formula, we will know how much volume of the 1.50 M we actually need.

1.50 M * V1 = 0.100 M * 350 mL

V1 = 23.33 mL

So we need 23.33 mL of the 1.50 M solution. We dilute it with water to a volume of 350 mL. So water needed is:

350 mL – 23.33 mL = 326.67 mL water

Steps:

1. Take 23.33 mL of 1.50 M solution

<span>2. Add 326.67 mL of water to make 350 mL of 0.100 M solution</span>

7 0

3 years ago

Give one example of Lewis acid

erma4kov [3.2K]

Answer:

<em><u>Copper </u></em><em><u>(</u></em><em><u>Cu2)</u></em><em><u> </u></em><em><u>,</u></em><em><u> </u></em><em><u>Iron </u></em><em><u>(</u></em><em><u>Fe2+</u></em><em><u> </u></em><em><u>Fe3 </u></em><em><u>+</u></em><em><u>)</u></em><em><u> </u></em><em><u>,</u></em><em><u> </u></em><em><u>and </u></em><em><u>Hydrogen </u></em><em><u>ion </u></em><em><u>(</u></em><em><u>H+</u></em><em><u>)</u></em>

Explanation:

I hope it helps u dear! ^_^

8 0

2 years ago

Is this a correct response? Please help me!

tia_tia [17]

Yes bsjaibsvdnakiebdnsoehdbdnsieveb

4 0

3 years ago

Read 2 more answers

How many moles of hydrogen are produced when 6.28 mil of oxygen form

sergejj [24]

If im correct i think the answer is 12.6 mol.

6 0

3 years ago