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Ghella [55]
2 years ago
12

Phenol (C6H5OH) is often used as an antiseptic in mouthwashes and throat lozenges. If a mouthwash has a phenol concentration of

1.5 g per 100 mL of solution, what is the molarity of phenol?
Chemistry
1 answer:
amm18122 years ago
3 0

Answer:

M=0.16M

Explanation:

Hello!

In this case, since the molarity is defined as moles of solute divided by liters of solution, since we have phenol with a molar mass of 94.12 g/mol, we can first compute the moles in 1.5 g of phenol:

n=1.5g*\frac{1mol}{94.12 g}=0.016mol

Next, since 1000 mL = 1 L, we notice that the volume of the solution is 0.100 L and therefore, the molarity of such solution turns out:

M=\frac{n}{V}=\frac{0.016mol}{0.100L}\\\\M=0.16M

Best regards!

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Zn + 2HCI --> ZnCl2 + H2
Elena-2011 [213]
Moles= mass\ relative formula mass(Ar)
moles of zinc= 7.9/30= 0.263
so we have 0.263 moles of zinc, and you need twice the amount of chlorine so therefore 0.526moles of chlorine= 0.526x 17=8.942g of chlorine
i cba to work the rest out but the most reasonable answer is 0.24 mol however if you need to use working outs, use the formula i provided earlier
8 0
2 years ago
The acid HOCl (hypochlorous acid) is produced by bubbling chlorine gas through a suspension of solid mercury(II) oxide particles
Oliga [24]

Answer:

K = [ HOCl ] . [HgO. HgCl2] / [Cl2]^2 [H2O] [HgO]^2

Explanation:

The law of Mass Action states that, at constant temperature, the rate of reaction is proportional to the active masses of each of the reactants.

The reaction above is a reversible reaction and the law of mass action also applies to it.

The rate of reaction from left-to-right reaction = r1 = k. [Cl2]^2 [H2O] [HgO]^2

Rate of reaction from right - to - left r2 = k. [hocl]^2 [HgO . hgcl2]

Then at equilibrium,

r1 = r2

k1/k2 = [HOCl ]^2 [HgO. HgCl2] / [Cl2]^2 [H2O] [HgO]^2 = K

where K is the equilibrium constant for the reaction.

5 0
3 years ago
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Answer:

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4 0
3 years ago
A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume bomb calorimeter. Consequently, the temperature rose b
dezoksy [38]

Answer:

872.28 kJ/mol

Explanation:

The heat released is:

ΔH = C*ΔT

where ΔH is the heat of combustion, C is the heat capacity of the bomb plus water, and ΔT is the rise of temperature. Replacing with data:

ΔH =  9.47*5.72 = 54.1684kJ

A quantity of 1.922 g of methanol in moles are:

moles = mass / molar mass

moles = 1.992/32.04 = 0.0621 mol

Then the molar heat of combustion of methanol is:

ΔH/moles = 54.1684/0.0621 = 872.28 kJ/mol

5 0
3 years ago
Ammonia and oxygen react to form nitrogen and water.
Nata [24]

Answer:

A. 19.2 g of O2.

B. 3.79 g of N2.

C. 54 g of H2O.

Explanation:

The balanced equation for the reaction is given below:

4NH3(g) + 3O2(g) → 2N2+ 6H2O(g)

Next, we shall determine the masses of NH3 and O2 that reacted and the masses of N2 and H2O produced from the balanced equation.

This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96 g

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 from the balanced equation = 2 x 28 = 56 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 6 x 18 = 108 g

Summary:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2 to produce 56 g of N2 and 108 g of H2O.

A. Determination of the mass of O2 needed to react with 13.6 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2.

Therefore, 13.6 g of NH3 will react with = (13.6 x 96)/68 = 19.2 g of O2.

Therefore, 19.2 g of O2 are needed for the reaction.

B. Determination of the mass of N2 produced when 6.50 g of O2 react.

This is illustrated below:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Therefore, 6.5 g of O2 will react to produce = (6.5 x 56)/96 = 3.79 g of N2.

Therefore, 3.79 g of N2 were produced from the reaction.

C. Determination of the mass of H2O formed from the reaction of 34 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted to 108 g of H2O.

Therefore, 34 g of NH3 will react to produce = (34 x 108)/68 = 54 g of H2O.

Therefore, 54 g of H2O were obtained from the reaction.

4 0
2 years ago
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