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3 years ago

1. Burning a candle is a chemical change. What is the primary evidence that burning the candle is a chemical reaction?

Chemistry

1 answer:

tankabanditka [31]3 years ago

8 0

<h2>Answer 1) Option B) Temperature Change</h2><h3>Explanation : </h3>

The burning of candle is the reaction where both physical and chemical changes occurs. A physical change where the melted wax is cooled and the wax is obtained again. Whereas the chemical change takes place when the wick of the candle burns using the oxygen in the air and produces heat and light. So, this change in temperature is a chemical change as the burning process cannot be reverse, from the burnt products like soot, smoke and carbon dioxide. Therefore, the temperature change is the evidence of a chemical change in the burning of candle.

<h2>Answer 2) Option B) It is a chemical change because heat energy is absorbed and the pack gets colder.</h2><h3>Explanation :</h3>

When Clara is mixing the vial contents the ice pack gets colder. On application of the ice pack on the injured area absorbs the heat from that region and becomes more colder. This happens because in the ice pack was the medium for the temperature change that was occurring. This temperature change by absorbing heat getting colder is the evidence of a chemical change.

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Look at the table. Which of these three substances is not an alloy?

Gemiola [76]

Answer:

B is the correct option.

5 0

2 years ago

Read 2 more answers

Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +

qwelly [4]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

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

The given chemical reaction follows:

$X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)$ $\Delta H_1=-668.5kJ$

(2) $XCO_3(s)\rightarrow XO(s)+CO_2$ $\Delta H_2=+384.3kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

7 0

3 years ago

grandma baked cookies. she gave 7 to sandy. she gave half of what were left, plus one to sue. she gave 3 to stan. then she gave

scoray [572]

I believe your answer is 23.
Credit: answers.yahoo.com

Hope this helps!

8 0

3 years ago

What is acid sulfuric's function ?

masya89 [10]

Answer:

The answer is below

Explanation:

Sulfuric acid is used for various purposes, some of which include the following:

1. It is used in the production of various manufactured goods.

2. It is used in the manufacturing of chemicals

3. It is also used in the making of fertilizer

4. It is used in the refining process of petroleum products

5. It is used in the processing of metals

3 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

marin [14]

Answer:

$\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}$

Explanation:

1. Calculate the moles of copper(II) hydroxide

$\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}$

2. Calculate the molecules of copper(II) hydroxide

$\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}$

6 0

3 years ago