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Masja [62]
3 years ago
9

What is the hydronium ion concentration in a solution prepared by mixing 150.00 ml of 0.10 mhcn with 150.00 ml of 0.030 mnacn? a

ssume that the volumes of the solutions are additive and that ka=4.9×10−10. what is the hydronium ion concentration in a solution prepared by mixing 150.00 of 0.10 with 150.00 of 0.030 ? assume that the volumes of the solutions are additive and that . 4.9×10−10m 7.0×10−6m 1.6×10−9m 2.4×1010m?
Chemistry
1 answer:
PtichkaEL [24]3 years ago
7 0
Answer is: <span>1.6×10−9m.
</span><span>This is buffer solution, so use Henderson–Hasselbalch equation: 
</span>V(solution) = 150 mL + 150 mL = 300 mL ÷ 1000 mL/L = 0.3 L.
c(HCN) = ck = 0.15 L · 0.1 M / 0.3 L = 0.05 M.
c(NaCN) = cs = 0.15 L · 0.03 M / 0.3 M = 0.015 M.
[H⁺] = Ka · (ck/cs).
[H⁺] = 4.9·10⁻¹⁰ · (0.05 M ÷ 0.015 M).
[H⁺] = 1.63·10⁻⁹ M.

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Explanation:

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using the Bohr model for hydrogen: energy = hc/wavelength = 2.18 x 10^-18 Joules (1/nf2 - 1/ni2) N=15 to n=5
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Answer:

Energy lost is 7.63×10⁻²⁰J

Explanation:

Hello,

I think what the question is requesting is to calculate the energy difference when an excited electron drops from N = 15 to N = 5

E = hc/λ(1/n₂² - 1/n₁²)

n₁ = 15

n₂ = 5

hc/λ = 2.18×10⁻¹⁸J (according to the data)

E = 2.18×10⁻¹⁸ (1/n₂² - 1/n₁²)

E = 2.18×10⁻¹⁸ (1/15² - 1/5²)

E = 2.18×10⁻¹⁸ ×(-0.035)

E = -7.63×10⁻²⁰J

The energy lost is 7.63×10⁻²⁰J

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Convert 4.50 e23 atoms of CO to grams.
hichkok12 [17]

Answer:

10.4664 grams of CO

Explanation:

Remark

There's a couple of things you must look out for in this question.

1. The use of the term atoms. There are 2 atoms in each mol of CO

2. You need to divide by 2 to find the number of molecules which will lead to moles.

<u>Step one</u>

Divide the number of atoms by 2

4.50 e^23 / 2 atoms = 2.25 * 10^23 molecules.

<u>Step Two</u>

Find the number of moles of CO

1 mol of anything is 6.02 * 10^23 molecules in this case

x  = 2.25 * 10^23 molecules

1/x = 6.02 * 10^23/2.25 * 10 ^23            Cross multiply

1 * 2.25 * 10^23 = 6.02*10^23 * x           Divide by 6.02 * 10^23

2.25 * 10 ^ 23 / 6.02 * 10^23 = x

x = .3738 moles of CO

<u>Step Three</u>

Find the gram molecular mass of CO

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1 mole = 12 + 16  = 28 grams.

<u>Step Four</u>

Find the number of gram in 0.3738 mols

1 mol = 28 grams

0.3738 mol = x                Cross multiply

x = 28 * 0.3738

x = 10.4664 grams

7 0
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