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saw5 [17]
3 years ago
10

How many moles of neon (Ne) gas have a volume of 0.84 L and a pressure of

Chemistry
1 answer:
kiruha [24]3 years ago
7 0
Maybe b I’m not sure
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Exactly how much time must elapse before 16 grams of potassium-42decays, leaving 2 grams of the original isotope?(1) 8 × 12.4 ho
AleksandrR [38]
The answer is <span>(3) 3 × 12.4 hours
</span>
To calculate this, we will use two equations:
(1/2) ^{n} =x
t_{1/2} = \frac{t}{n}
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>t_{1/2} - half-life length
</span>t - total time elapsed.

First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125

Thus:
<span>(1/2) ^{n} =x
</span>(0.5) ^{n} =0.125
n*log(0.5)=log(0.125)
n= \frac{log(0.5)}{log(0.125)}
n=3

It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>t_{1/2} = 12.4
</span><span>t_{1/2} *n = t
</span>t= 12.4*3

Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>
7 0
3 years ago
Read 2 more answers
Is sublimation exothermic or endothermic
Masja [62]

Answer:

endormic

Explanation:

It occurs at a temperature and pressures below a substance's triple point on its phase diagram, which corresponds to the lowest pressure at which the substance can exist as a liquid

5 0
2 years ago
Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find
FromTheMoon [43]

<u>Answer:</u> The concentration of SO_4^{2-} at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

       H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)

            0.025          0.025       0.025

Equation for the second dissociation of sulfuric acid:

                    HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)

<u>Initial:</u>            0.025            0.025      

<u>At eqllm:</u>      0.025-x          0.025+x        x

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

We know that:

Ka_2\text{ for }H_2SO_4=0.01

Putting values in above equation, we get:

0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of SO_4^{2-} at equilibrium is 0.00608 M

4 0
3 years ago
Balance the following equation. __AI+ __CuO __AI203+ _Cu
Nesterboy [21]
2 Al+ 3 CuO-> 1 Al2O3+ 3Cu
7 0
2 years ago
What are the 3 forces that make the tectonic plates<br> move
Molodets [167]

Answer:

Viscous Drag.

Slab-Pull Force.

Ridge-Push Force

Explanation:

your welcome have a nice day

8 0
3 years ago
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