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3 years ago

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An element has two isotopes. 90% of the isotopes have a mass number of 20 amu, while 10% have a mass number of 22 amu. Calculate

the Average Atomic Mass of the element

1 answer:

jek_recluse [69]3 years ago

8 0

Explanation:

aam= 90%•20amu+10%•22amu/100

aam= 2020/100

aam= 20.2

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How many cm3 are there in 0.25 dm3?

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Read 2 more answers

Calculate the pH of 1.00 L of the buffer 0.95 M CH3COONa/0.92 M CH3COOH before and after the addition of the following species.

natima [27]

Answer: (a) pH = 4.774, (b) pH = 4.811 and (c) pH = 4.681

Explanation: (a) pH of the buffer solution is calculated using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

pKa for acetic acid is 4.76. concentration of base and acid are given as 0.95M and 0.92M. Let's plug in the values in the equation and calculate the pH of starting buffer.

$pH=4.76+log(\frac{0.95}{0.92})$

pH = 4.76 + 0.014

pH = 4.774

(b) When 0.040 mol of NaOH (strong base) are added to the buffer then it reacts with 0.040 mol of acetic acid and form 0.040 mol of sodium acetate.

Original buffer volume is 1.00 L. So, the original moles of sodium acetate will be 0.95 and acetic acid will be 0.92.

moles of acetic acid after addition of NaOH = 0.92 - 0.040 = 0.88

moles of sodium acetate after addition of NaOH = 0.95 + 0.040 = 0.99

Let's again plug in the values in the Handerson equation:

$pH=4.76+log(\frac{0.99}{0.88})$

pH = 4.76 + 0.051

pH = 4.811

(c) When 0.100 mol of HCl are added then it reacts with exactly 0.100 moles of sodium acetate(base) and form 0.100 moles of acetic acid(acid).

so, new moles of acetic acid = 0.92 + 0.100 = 1.02

new moles of sodium acetate = 0.95 - 0.100 = 0.85

Let's plug in the values in the equation:

$pH=4.76+log(\frac{0.85}{1.02})$

pH = 4.76 - 0.079

pH = 4.681

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4 years ago

Which of the following is an example of degradation?

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3 years ago

The vapor pressure of ethanol is 400 mmhg at 63.5°c. its molar heat of vaporization is 39.3 kj/mol. what is vapor pressure of et

SVETLANKA909090 [29]

Answer:- The pressure of ethanol would be 109 mmHg.

Solution:- This problem is based on Clausius clapeyron equation--

$ln(\frac{P_1}{P_2})=(\frac{\Delta Hvap}{R})(\frac{1}{T_2}-\frac{1}{T_1})$

Given, $T_1$ = 63.5 + 273 = 336.5 K

$T_2$ = 34.9 + 273 = 307.9 K

$P_1$ = 400 mmHg

$P_2$ = ?

$\Delta Hvap$ = 39.3 kJ/mol = 39300 J/mol

R = 8.314 J/mol.K

Let's plug in the values in the equation and do the calculations.

$ln(\frac{400}{P_2})=(\frac{39300}{8.314})(\frac{1}{307.9}-\frac{1}{336.5})$

$ln(\frac{400}{P_2})$ = 1.30

On taking anti ln to both sides...

$\frac{400}{P_2}$ = $e^1^.^3^0$

$\frac{400}{P_2}$ = 3.67

$P_2$ = 400/3.67

$P_2$ = 109 mmHg

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3 years ago