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yaroslaw [1]
3 years ago
11

Rn oxidation number is... -3 -2 0 +1

Chemistry
1 answer:
eimsori [14]3 years ago
6 0
I’m pretty sure it’s -2 my friend said it ‍♂️
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A balloon, whose volume at 23°C is 535 mL, is heated to 46°C. Assuming the pressure and amount of gas remain constant, what is t
NikAS [45]
This problem requires a certain equation.  That equation is V1/T1=V2/T2, where V1 is your initial volume (535 mL in this case), T1 is your initial temperature in Kelvin(23 degrees C = 296 K), V2 is your final volume (unknown), and T2 is your final temperature (46 degrees C = 319 K). By plugging in these values, the equation looks like this: 535/296=V2/319.  Now multiply both sides of the equation by 319, and your final answer is V2= 576.6 mL
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[H_{3}O^{+}] = 0.00770 M

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Given [H_{3}O^{+}] = 0.00770 M

Let the initial concentration of acid be x and change y

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K_{a} = \frac{(0.00770 M)(0.00770 M)}{x - 0.00770}

0.00257 = \frac{0.00005929}{x - 0.00770}

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Therefore, initial concentration of the weak acid is <u>0.031 M</u>

4 0
4 years ago
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trasher [3.6K]
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2. quickest

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3 0
3 years ago
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