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3 years ago

If a decrease in temperature accompanies a reaction, what occurred?

Chemistry

1 answer:

Alina [70]3 years ago

6 0

<h2>Answer:</h2>

If a decrease in temperature accompanies a reaction, Energy was absorbed.

So the forth option is correct option.

<h3>Explanation:</h3>

If decrease in temperature occurs during a reaction, it indicates that the reaction is endothermic reaction.
<u>Endothermic process describes the process or reaction in which the system absorbs energy from its surroundings, usually in the form of heat.</u>
<em>So energy in form of heat is absorbed.</em>

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A) Cl2 + 2NaI = 2NaCl + I2

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PLEASE HELP!!!!!!!!Which of the following chemical reactions is best clasified as an acid-base reaction?

ANTONII [103]

Answer:

d . H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

Explanation:

The given reactions are:

a . 2HgO → 2Hg + O₂,

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b . Na₂SO₄ + BaCl₂ → BaSO₄ + 2NaCl,

It is a double replacement reaction where two salts replaces their cations and anions with each others producing 2 new salts.

c . Zn + 2AgNO₃ → Zn(NO₃)₂ + 2Ag,

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4 years ago

Determine the limiting reactant when 30.0 g of propane, C3H8, is burned with 75.0 g of oxygen.

AnnZ [28]

<u>Answer:</u> The limiting reagent is oxygen gas.

<u>Explanation:</u>

Limiting reagent is defined as the reactant that is present in less amount and it limits the formation of products.

Excess reagent is defined as the reactant which is present in large amount.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For propane:</u>

Given mass of propane = 30.0 g

Molar mass of propane = 44.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of propane}=\frac{30.0g}{44.1g/mol}=0.680mol$

<u>For oxygen:</u>

Given mass of oxygen = 75.0 g

Molar mass of oxygen = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of propane}=\frac{75.0g}{32g/mol}=2.34mol$

The chemical equation for the combustion of propane follows:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)$

By Stoichiometry of the reaction:

5 moles of oxygen gas reacts with 1 mole of propane.

So, 2.34 moles of oxygen gas will react with = $\frac{1}{5}\times 2.34=0.468mol$ of propane

As, given amount of propane is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reagent is oxygen gas.

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3 years ago

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Answer:

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Explanation:

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