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2 years ago

Which of the following best describes a mole as used in chemistry?

Chemistry

1 answer:

bazaltina [42]2 years ago

8 0

Answer: a very large number of objects

Explanation: Mole is the amount of objects and it has the usually Avogadro number of atoms, molecules, ions, etc in chemistry

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There was a $13 difference

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2 years ago

Describe briefly the changes from the control values in the MEAN values for volume, specific gravity and NaCl after drinking pur

Yuri [45]

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3 years ago

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

3 0

2 years ago

Determine the number of grams of Carbon in 215 g of sucrose.

olya-2409 [2.1K]

<span>just find the percent mass of oxygen in sucrose again. and then multiply that by 50.00.</span>

6 0

3 years ago

What volume. In liters, of H2O(g) measured at STP is produced by the combustion of 15.63 g of natural gas (CH4) according to the

fomenos

Answer:

V = 43.95 L

Explanation:

Given data:

Mass of CH₄ decomposed = 15.63 g

Volume of H₂O produced at STP = ?

Solution:

Chemical equation:

CH₄ + 2O₂ → 2H₂O + CO₂

Number of moles of CH₄:

Number of moles = mass/molar mass

Number of moles = 15.63 g/ 16 g/mol

Number of moles = 0.98 mol

Now we will compare the moles of H₂O with CH₄.

CH₄ : H₂O

1 : 2

0.98 : 2×0.98 = 1.96 mol

Volume of hydrogen:

PV = nRT

1 atm × V = 1.96 mol × 0.0821 atm.L/mol.K × 273.15 K

V = 43.95atm.L / 1atm

V = 43.95 L

3 0

2 years ago