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3 years ago

If the stack of bricks on the right has a mass of 240 kg, approximately what is the mass of the stack of bricks on the left?

Chemistry

2 answers:

solong [7]3 years ago

4 0

Answer:

720

Explanation: study island

Vikki [24]3 years ago

3 0

Answer:

720

Explanation:

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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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Automobile bodies contain significant amounts of iron. The iron is protected by the addition of zinc. This is called galvanizati

Oksana_A [137]

<u>Answer:</u> The balanced chemical equation is written below.

<u>Explanation:</u>

Galvanization is defined as the process in which a protective layer of zinc is applied to iron or steel to prevent the metal from rusting.

Zinc prevents the oxidation of iron and acts as a reducing agent in the process.

The half reaction for the process follows:

<u>Oxidation half reaction:</u> $Zn\rightarrow Zn^{2+}+2e^-$

<u>Reduction half reaction:</u> $Fe^{2+}+2e^-\rightarrow Fe$

Net chemical equation: $Zn+Fe^{2+}\rightarrow Zn^{2+}+Fe$

Hence, the balanced chemical equation is written above.

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