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Rudik [331]
3 years ago
11

Which of these contains generic material but is not classified as living?

Chemistry
1 answer:
denis-greek [22]3 years ago
4 0
Viruses are not classified as being alive in because of the fact that they can't reproduce.
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What is the temperature of 0.47 mol of gas at a pressure of 1.5 atm and a volume of 10.5 l ?
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The temp is 0.002448 of the equation
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The liter is used to measure _____.<br><br> weight<br> mass<br> volume<br> length
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the answer to your question is

volume

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Explanation:

d. endothermic change as

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3 years ago
The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. Calculate the equilibrium concentration of H3O+ in the s
raketka [301]

Answer : The equilibrium concentration of H_3O^+ in the solution is, 2.1\times 10^{-3}M

Explanation :

The dissociation of acid reaction is:

                       C_6H_5COOH+H_2O\rightarrow H_3O^++C_6H_5COO^-

Initial conc.        c                                 0                0

At eqm.             c-x                                 x                x

Given:

c = 7.0\times 10^{-2}M

K_a=6.3\times 10^{-5}

The expression of dissociation constant of acid is:

K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}

K_a=\frac{(x)\times (x)}{(c-x)}

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

6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}

x=2.1\times 10^{-3}M

Thus, the equilibrium concentration of H_3O^+ in the solution is, 2.1\times 10^{-3}M

4 0
3 years ago
Find the empirical formula of the compound ribose, a simple sugar often used as a nutritional supplement. A 14.229 g sample of r
MakcuM [25]

Answer:

CH2O

Explanation:

Firstly, we need to convert the masses of the elements to percentage compositions. This can be done by placing the mass of each element over the total mass multiplied by 100% . We can start with carbon.

C = 5.692/14.229 * 100 = 40%

O = 7.582/14.229 * 100 = 53.29%

H = 0.955/14.229 * 100 = 6.71%

We then proceed to divide each percentage composition by their atomic mass of 12, 16 and 1 respectively.

C = 40/12 = 3.333

O = 53.29/16 = 3.33

H = 6.71/2 = 6.71

Dividing by the smaller value which is 3.33

C = 3.33/3.33 = 1

O = 3.33/3.33= 1

H = 6.71/3.33 = 2

The empirical formula of the compound ribose is CH2O

6 0
3 years ago
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