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leonid [27]
3 years ago
6

What is the density of a silver coin from the following data:

Chemistry
1 answer:
serious [3.7K]3 years ago
6 0

Answer:

you can use math wa, it helps

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Choose all the answers that apply.
Stolb23 [73]

Answer:

the first statement

Explanation:

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3 years ago
If the half-life of a sample of a radioactive substance is 30 seconds, how much would be left after 60 seconds? A. one-fourth B.
Basile [38]
m=m_{0}*(\frac{1}{2})^{\frac{60}{30}}\\\\
m=m_{0}*(\frac{1}{2})^{2}\\\\
m=\frac{1}{4}m_{0}

If the half-life of a sample of a radioactive substance is 30 seconds, how much would be left after 60 seconds? <span>A. one-fourth</span>
7 0
3 years ago
Read 2 more answers
Phenolphthalein has a pKa of 9.7 and is colorless in its acid form and pink in its basic form. calculate [In-}/{HIn} for the fol
zvonat [6]

Answer:

1.58x10⁻⁵

2.51x10⁻⁸

0.0126

63.10

Explanation:

Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:

pH = pKa + log[In-]/[HIn]

pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,

i) pH = 4.9

4.9 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = - 4.8

[In-]/[HIn] = 10^{-4.8}

[In-]/[HIn] = 1.58x10⁻⁵

ii) pH = 2.1

2.1 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -7.6

[In-]/[HIn] = 10^{-7.6}

[In-]/[HIn] = 2.51x10⁻⁸

iii) pH = 7.8

7.8 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -1.9

[In-]/[HIn] = 10^{-1.9}

[In-]/[HIn] = 0.0126

iv) pH = 11.5

11.5 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = 1.8

[In-]/[HIn] = 10^{1.8}

[In-]/[HIn] = 63.10

6 0
3 years ago
An acid with the general formula RCOOH is used to make a 0.10 M solution in water. The pH of this acid solution is measured as 3
Sati [7]

Answer:

0.159 \%

Explanation:

The acid will dissociate according to the reaction shown below:-

RCOOH + H_2O\rightleftharpoons RCOO^- + H_3O^+

Given that, pH=3.8

The concentration of can be determined from the expression fo pH as:-

pH = - log [H^+]

3.8  = - log [H^+]

[H^+] = 1.59\times 10^{-4}\ M

The initial concentration of RCOOH was 0.10 M, then the percent dissociation was- calculated as shown below:-

\% \ dissociation=\frac{1.59\times 10^{-4}}{0.10}\times 100=0.159\ \%

3 0
3 years ago
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