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3 years ago

Convert 321 millimeters to picometers

Chemistry

1 answer:

pantera1 [17]3 years ago

3 0

3.21e+11 is your answer for 321 millimeters in picometers. Hope this helps!

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HI(aq)+NaOH(aq)→ <br> what the final balanced chemical equation with the phases included

julia-pushkina [17]

Answer:

$HI(aq)+NaOH(aq)\rightarrow NaI(aq)+H_2O(l)$

Explanation:

Hello there!

In this case, for this neutralization reaction, it is possible to realize that one the neutralization products is water (pH=7) and the other one is the salt coming up from the cation of the NaOH and the anion of the HI:

$HI(aq)+NaOH(aq)\rightarrow NaI+H_2O$

Moreover, since the solubility of NaI is large in water, we infer it remains aqueous whereas the water is maintained as liquid:

$HI(aq)+NaOH(aq)\rightarrow NaI(aq)+H_2O(l)$

Which is also balanced as the number of atoms of all the elements is the same at both sides.

Best regards!

7 0

3 years ago

Part A

KonstantinChe [14]

Answer:

The answer to your question is below

Explanation:

[H₃O⁺] = 2 x 10⁻¹⁴ M

pH = ?

Formula

pH = - log [H₃O⁺]

Substitution

pH = - log [2 x 10⁻¹⁴]

Result

pH = 13.7

[H₃O⁺] = ?

pH = 3.12

Formula

pH = - log [H₃O⁺]

Substitution

3.12 = - log [H₃O⁺]

$10^{-3.12} = [H_{3} O^{+}]$

Result

[H₃O⁺] = 7.59 M

7 0

3 years ago

Michelle is trying to find the average atomic mass of a sample of an unknown

GREYUIT [131]

The average atomic mass of her sample is 114.54 amu

Let the 1st isotope be A

Let the 2nd isotope be B

From the question given above, the following data were obtained:

Abundance of isotope A (A%) = 59.34%
Mass of isotope A = 113.6459 amu
Mass of isotope B = 115.8488 amu
Abundance of isotope B (B%) = 100 – 59.34 = 40.66%
Average atomic mass =?

The average atomic mass of the sample can be obtained as follow:

$Average \: atomic \: mass \: = \frac{mass \: of \: A \times A\%}{100} + \frac{mass \: of \: B \times B\%}{100} \\ \\ Average \: atomic \: mass \: = \frac{113.6459\times 59.34}{100} + \frac{115.8488\times 40.66}{100} \\ \\ Average \: atomic \: mass \: = 114.54 \: amu \\ \\$