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AysviL [449]
3 years ago
11

A 6.68-g aqueous solution of isopropyl alcohol contains 3.97 g of isopropyl alcohol. What is the mass percentage of isopropyl al

cohol in the solution?
Chemistry
1 answer:
Sphinxa [80]3 years ago
6 0

Answer : The mass percentage of isopropyl alcohol in the solution is, 59.4 %

Explanation :

To calculate the mass percentage of isopropyl alcohol in the solution, we use the equation:

\text{Mass percent of isopropyl alcohol}=\frac{\text{Mass of isopropyl alcohol}}{\text{Mass of isopropyl alcohol solution}}\times 100

Mass of isopropyl alcohol = 6.68 g

Mass of isopropyl alcohol solution = 3.97 g

Putting values in above equation, we get:

\text{Mass percent of isopropyl alcohol}=\frac{3.97g}{6.68g}\times 100=59.4\%

Thus, the mass percentage of isopropyl alcohol in the solution is, 59.4 %

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A football is inflated to 12psi at 20°C What is the new pressure in the football
Radda [10]

Answer: The new pressure is 3 psi.

Explanation:

Given: P_{1} = 12 psi,         T_{1} = 20^{o}C

P_{2} = ?,          T_{2} = 5^{o}C

Formula used is as follows.

\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\\frac{12 psi}{20^{o}C} = \frac{P_{2}}{5^{o}C}\\P_{2} = \frac{12 psi \times 5^{o}C}{20^{o}C}\\= 3 psi

Thus, we can conclude that the new pressure is 3 psi.

7 0
3 years ago
1/2 - 3/7=??????????????????????​
Veseljchak [2.6K]
1/2 - 3/7 = 7/14 - 6/14 = 1/14
6 0
2 years ago
What would be the correct order of least(floats) to most(sinks) density?*
olya-2409 [2.1K]

Answer:

B

Explanation:

Water-Continental-Oceanic-Mantle

5 0
2 years ago
Calculate the vapor pressure in torr of a solution containng 24.5 g of glycerin (C3H8O3) in 135 mL water at 30.0* C; the vapor p
Otrada [13]
Psolution = X · PH_20
= 0.966 · 31.8 torr
= 30.7 torr
3 0
2 years ago
The salt formed by the reaction of the weak acid hydrocyanic acid, HCN, with the strong base potassium hydroxide is
iogann1982 [59]

Answer:

2.28 × 10^-3 mol/L

Explanation:

The equation for the equilibrium is

CN^- + H2O ⇌ HCN + OH^-

                    Ka = 4.9 × 10^-10

               KaKb = Kw

4.9 × 10^-10 Kb = 1.00 × 10^-14

                   Kb = (1.00 × 10^-14)/(4.9 × 10^-10) = 2.05 × 10^-5

Now, we can set up an ICE table

                     CN^- + H2O ⇌ HCN + OH^-

I/(mol/L)      0.255                     0         0

C/(mol/L)       -x                        +x        +x

E/(mol/L)  0.255 - x                   x         x

Ka = x^2/(0.255 - x) = 2.05 × 10^-5

Check for negligibility

0.255/(2.05 × 10^-5) = 12 000 > 400. ∴ x ≪ 0.255

    x^2 = 0.255(2.05 × 10^-5) = 5.20 × 10^-6

        x = sqrt(5.20 × 10^-6)    = 2.28 × 10^-3

[OH^-] = x mol/L                     = 2.28 × 10^-3 mol/L

5 0
2 years ago
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