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Gennadij [26K]
2 years ago
12

Points + Brainalist

Chemistry
1 answer:
Lorico [155]2 years ago
4 0

Answer:

I think A it looks to be the answer

Explanation:

sorry if wrong

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10. When the pressure on a gas inetcases three times, by how much will the volume incrcase or decrease?
blagie [28]

Answer:The answer to this question comes from experiments done by the scientist Robert Boyle in an effort to improve air pumps. In the 1600's, Boyle measured the volumes of gases at different pressures. Boyle found that when the pressure of gas at a constant temperature is increased, the volume of the gas decreases. when the pressure of gas is decreased, the volume increases. this relationship between pressure and volume is called Boyle's law.

Explanation: So, at constant temperature, the answer to your answer is: the volume decreases in the same ratio as the ratio of pressure increases.

BUT, in general, there is not a single answer to your question. It depend by the context.

For example, if you put the gas in a rigid steel tank (volume is constant), you can heat the gas, so provoking a pressure increase. But you won't get any change in volume.

Or, if you heat the gas in a partially elastic vessel (as a tire or a soccer ball) you will get both an increase of volume AND an increase of pressure.

FINALLY if you inflate a bubblegum ball, the volume will be increased without any change in pressure and temperature, because you have increased the NUMBER of molecules in the balloon.

There are many other ways to change volume and pressure of a gas that are different from the Boyle experiment.

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3 years ago
What does this motion cause on the surface of the earth (rotation)
vichka [17]
The motion causes day and night because the earth is constantly spinning on it's axis.
3 0
3 years ago
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How to convert 12g of H2O to moles
Anna35 [415]

Answer:

moles = given mass/atomic mass

so H2O mass = 2 +16=18

so 12g of h2o= 12/16 = 3/4 moles

8 0
2 years ago
A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci
Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}

The ICE table is then shown as:

                               C_3H_6ClCO_2H_{(aq)}  \ \ \ \ \to  \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ +  \ \ \ \ H^+_{(aq)}

Initial   (M)                     1.8                                       0                               0

Change  (M)                   - x                                     + x                           + x

Equilibrium   (M)            (1.8 -x)                                  x                              x

K_a  = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}

where ;

K_a = 3.02*10^{-5}

3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}

Since the value for K_a is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

3.02*10^{-5} *(1.8) = {(x)(x)}

5.436*10^{-5}= {(x^2)

x = \sqrt{5.436*10^{-5}}

x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M

Dissociated form of  4-chlorobutanoic acid = C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M

Percentage dissociated = \frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100

Percentage dissociated = \frac{7.3729*10^{-3}}{1.8 }*100

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

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