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2 years ago

24. A laboratory chemist wants to produce 25.0 g NO2 by the following reactions. If the

Chemistry

1 answer:

lyudmila [28]2 years ago

5 0

The amount of N2O5 to start with would be 35.79 grams

<h3>Stoichiometric calculations</h3>

From the balanced equation of the reactions:

$2N_2O_5 --- > 4NO_2 + O_2$

Mole ratio of N2O5 and NO2 = 1:2

Since the reaction's actual yield is 82%, the theoretical yield would be: 25 x 100/82 = 30.49 grams

Mole of 25 g NO2 = 30.49/46

= 0.66 mole

Equivalent mole of N2O5 = 0.54/2 mole

= 0.33 moles

Mass of 0.27 mole N2O5 = 0.33 x 108.01

= 35.79 grams

More on stoichiometric calculations can be found here: brainly.com/question/8062886

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Could someone explain how they got this answer, explain step by step plz

gulaghasi [49]

Answer:

6.018 amu

Explanation:

Let 6–Li be isotope A.

Let 7–Li be isotope B.

Let the abundance of 6–Li be A%

Let the abundance of 7–Li be B%

The following data were obtained from the question:

Atomic mass of isotope A (6–Li) =.?

Atomic mass of isotope B (7–Li ) = 7.015 amu.

Abundance of 7–Li (B%) = 92.58%

Abundance of 6–Li (A%) = 100 – B% = 100 – 92.58 = 7.42%

Atomic mass of Lithium = 6.941amu

The atomic mass of isotope A (6–Li) can be obtained as follow:

Atomic mass = [(Mass of A x A%)/100] + [(Mass of B x B%)/100]

6.941 = [(mass of A x 7.42)/100] + [(7.015x92.58)/100]

6.941 = [(mass of A x 7.42)/100] + 6.494487

(mass of A x 7.42)/100 = 6.941 – 6.494487

(mass of A x 7.42)/100 = 0.446513

Mass of A x 7.42 = 100 x 0.446513

Mass of A x 7.42 = 44.6513

Divide both side by 7.42

Mass of A = 44.6513 / 7.42

Mass of A = 6.018 amu

Therefore, the mass of 6–Li is 6.018 amu

7 0

3 years ago

Can someone plz explain ionic and covalent bonding plzzzz!!!

Alexeev081 [22]

Answer:

Ionic Bonding: The formation of an Ionic bond is the result of the transfer of one or more electrons from a metal onto a non-metal.

Covalent Bonding: Bonding between non-metals consists of two electrons shared between two atoms.

Explanation:

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3 years ago

What historical event of commercial interest triggered investigation in the science of steam?

Cerrena [4.2K]

Answer: Photosynthesis

Explanation:

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3 years ago

What layer of earths atmosphere is the layer next to the earth

Pavlova-9 [17]

Answer:

troposphere

Explanation:

5 0

3 years ago

A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL of 0.150 M NiSO4. (a) Write the balanced chemical equ

NISA [10]

Answer:

a) 2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂

b) Ni(OH)₂

c) KOH

d) 0.927 g

e) K⁺=0.067 M, SO₄²⁻=0.1 M, Ni²⁺=0.067 M

Explanation:

a) The equation is:

2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂ (1)

b) The precipitate formed is Ni(OH)₂

c) The limiting reactant is:

$n_{KOH} = V*M = 100.0 \cdot 10^{-3} L*0.200 mol/L = 0.020 moles$

$n_{NiSO_{4}} = V*M = 200.0 \cdot 10^{-3} L*0.150 mol/L = 0.030 moles$

From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:

$n = \frac{2}{1}*0.030 moles = 0.060 moles$

Hence, the limiting reactant is KOH.

d) The mass of the precipitate formed is:

$n_{Ni(OH)_{2}} = \frac{1}{2}*n_{KOH} = \frac{1}{2}*0.020 moles = 0.010 moles$

$m = n*M = 0.010 moles*92.72 g/mol = 0.927 g$

e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:

$C_{K^{+}} = \frac{2*\frac{1}{2}*n_{KOH}}{V} = \frac{0.020 moles}{0.300 L} = 0.067 M$

$C_{SO_{4}^{2-}} = \frac{\frac{1}{2}*n_{KOH + (0.03 - 0.01)}}{V} = \frac{0.030 moles}{0.300 L} = 0.1 M$

$C_{Ni^{2+}} = \frac{0.020 moles}{0.300 L} = 0.067 M$

I hope it helps you!

5 0

3 years ago