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2 years ago

How many grams of nicotine (C10H14N2) are in a 0.125 mol sample of nicotine?

1 answer:

Ivan2 years ago

5 0

Taking into account the molar mass of the compound, 20.25 grams of nicotine (C₁₀H₁₄N₂) are in a 0.125 mol sample of nicotine.

In first place, you have to know that molar mass is a physical property that is defined as the mass of a mole of a substance, which can be an element or a compound.

In a compound the molar mass is equal to the sum of the weight or atomic mass of its atoms multiplied by the quantity of each atom.

In the periodic table, it is possible to find the molar mass of the elements, also called atomic mass or atomic weight.

The molar mass of the elements, in this case, are:

C= 12 g/mole
H= 1 g/mole
N= 14 g/mole

So, the molar mass of the compound is:

C₁₀H₁₄N₂= 10× (12 g/mole) + 14× (1 g/mole) + 2× (14 g/mole)

Then you can apply the following rule of three: if 1 mole of nicotine contains 162 g of the compound, 0.125 moles contains how much mass?

$mass=\frac{0.125 molesx162 grams}{1 mole}$

Solving:

mass= 20.25 grams

In summary, 20.25 grams of nicotine (C₁₀H₁₄N₂) are in a 0.125 mol sample of nicotine.

Learn more about molar mass:

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How many moles of oxygen are formed when 58.6 g of KNO3 decomposes according to the following reaction? 4 KNO3(s) → 2 K2O(s) + 2

Leya [2.2K]

Answer:

0.725 mol

Explanation:

Moles are calculated as the given mass divided by the molecular mass.

i.e. ,

moles = ( mass / molecular mass )

since,

mass of KNO₃ = 58.6 g ( given )

Molecular mass of KNO₃ = 101 g / mol

Therefore,

moles of KNO₃ = 58.6 g / 101 g / mol

moles of KNO₃ = 0.58 mol

From the balanced reaction ,

4 KNO₃ (s) ---> 2K₂O (s) + 2N₂ (g) + 5O₂ (g)

By the decomposition of 4 mol of KNO₃ , 5 mol of O₂ are formed ,

hence, unitary method is used as,

1 mol of KNO₃ gives 5 / 4 mol O₂

Therefore,

0.58 mol of KNO₃ , gives , 5 / 4 * 0.58 mol of O₂

Solving,

0.58 mol of KNO₃ , gives , 0.725 mol of O₂

Therefore,

58.6g of KNO₃ gives 0.725 mol of O₂.

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Explanation:

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Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The Ks

charle [14.2K]

Answer:

pH = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH- Ksp = 3X10^-15

Zn+2 + 4OH- --> Zn(OH)4-2 Kf = 2X10^15

K = Ksp X Kf

= 3*2*10^-15 * 10^15

= 6

Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M

Zn(OH)₂ + 2OH⁻(aq) --> Zn(OH)₄²⁻(aq)

Initial: 0 0.3 0

Change: -2x +x

Equilibrium: 0.3 - 2x x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

= 0.3 -(2*0.089)

= 0.122

pOH = -log[OH⁻]

= -log 0.122

= 0.91

pH = 14-0.91

= 13.09

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3 years ago

Net ionic equation for Barium chloride

vaieri [72.5K]

Answer: There is one way to write it but i’ll also provide an unbalanced equation and a balanced one.

Explanation:

Unbalanced : Ba (aq) + Cl2 (aq)—-> BaCl (aq)

Balanced : 2Ba (aq) + Cl2 (aq)—> 2BaCl(aq)

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3 years ago