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3 years ago

17. Which is a fat-soluble vitamin? O vitamin B O vitamin C O vitamin A

Chemistry

1 answer:

vichka [17]3 years ago

4 0

Vitamin A is the fat soluble

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Read the scientific question below.

svetlana [45]

If we fertilize a plant, then its height increases fast. Always use if then format

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3 years ago

Which best describes a compound such as sodium chloride?

gayaneshka [121]

Answer:

check it below

Explanation:

NaCl; Sodium Chloride is an ionic compound formed by sodium and Chlorine.

Ionic bond is very strong, It can't be separated back to sodium and chlorine just by physical change. Chemicals which are more reactive can displace ions, thus seperate it

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3 years ago

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

Vilka [71]

Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

$K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}$

$K_c=1.066\aprrox 1.1$

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0

3 years ago

Light bending as it passes through a rain drop is an example of Transmission Diffraction Refraction Reflection

Yuliya22 [10]

The answer would be the third one listed, Refraction

4 0

3 years ago

What is the [OH-] of a substance that has a pH of 11?

Deffense [45]

Answer:

0.001 M OH-

Explanation:

[OH-] = 10^-pOH, so

pOH + pH = 14 and 14 - pH = pOH

14 - 11 = 3

[OH⁻] = 10⁻³ ; [OH-] = 0.001 M OH-

6 0

3 years ago