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mr Goodwill [35]
2 years ago
14

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]

Answer:

As specific gravity is defined as weight per unit volume,so urine is the less dense fluid,so urine will be in hyperosmotic category.

8 0
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
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