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frozen [14]
3 years ago
11

Which types of reactions would result in a change in temperature?

Chemistry
1 answer:
Law Incorporation [45]3 years ago
6 0

Answer:

Chemical reaction involves the breaking of bonds in the reactants and formation of bonds in the products. ... If a reaction is exothermic, more energy is released when the bonds of the products are formed than it takes to break the bonds of the reactants. This is the reason for temperature change during a reaction.

Explanation:

Here are just a few everyday demonstrations that temperature changes the rate of chemical reaction: Cookies bake faster at higher temperatures. Bread dough rises more quickly in a warm place than in a cool one.

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7 f Find the volume in dm3 and in mole of 0.505m of NaoH required to react with 40ml of 0.505m
Anna [14]

The volume of NaOH required is 0.08 dm³

To solve this question, we'll begin by writing the balanced equation for the reaction between H₂SO₄ and NaOH. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ (n_{A}) = 1

Mole ratio of the base, NaOH (n_{B}) = 2

Next, we shall determine the volume of NaOH required to react with H₂SO₄. This can be obtained as follow:

Molarity of the base, NaOH (M_{B}) = 0.505 M

Volume of the acid, H₂SO₄ (V_{A}) = 40 mL

Molarity of the acid, H₂SO₄ (M_{A}) = 0.505 M

<h3>Volume of the base, NaOH (V_{B}) =? </h3>

\frac{M_{A} * V_{A}}{M_{B} * V_{B}} = \frac{n_{A}}{n_{B}}\\\\\frac{0.505 * 40}{0.505 *V_{B}} = \frac{1}{2}\\\\\frac{20.2}{0.505 *V_{B}} = \frac{1}{2}

Cross multiply

0.505 * V_{B} = 20.2 * 2\\0.505 * V_{B} = 40.4

Divide both side by 0.505

V_{B} = \frac{40.4}{0.505}\\\\V_{B} = 80 mL

Finally, we shall convert 80 mL to dm³. This can be obtained as follow:

1000 mL = 1 dm^{3}\\\\Therefore,\\\\80 mL = \frac{80 mL * 1dm^{3}}{1000 mL}\\\\80 mL = 0.08dm^{3}

Therefore, the volume of NaOH required is 0.08 dm³

Learn more: brainly.com/question/19053582

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Chuẩn Độ 15 ml dung dịch CH3COOH 0,2 m bằng dung dịch NaOH 0,2 m
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Answer:

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3 0
3 years ago
In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2
valentinak56 [21]

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

\dfrac{80}{17}=4.706

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

\tt \dfrac{120}{32}=3.75

Mol ratio of reactants(to find limiting reatants) :

\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

\tt \dfrac{6}{5}\times 3.75=4.5

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%

6 0
3 years ago
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