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hodyreva [135]
3 years ago
6

A sample of gas (1.9 mol) is in a flask at 21 °C and 697 mmHg. The flask is opened and more gas is added to the flask. The new p

ressure is 795 mmHg and the temperature is now 26 °C. There are now __________ mol of gas in the flask.
Chemistry
1 answer:
mariarad [96]3 years ago
4 0

Answer:

The new moles of the gas in the flask is 2.13 moles.

Explanation:

Given;

number of moles of gas, n = 1.9 mol

temperature of the gas, T = 21 °C  = 21 + 273 = 294 K

pressure of gas, P = 697 mmHg

volume of gas, V = ?

Apply ideal gas law;

PV = nRT

Where;

R is gas constant, = 62.363 mmHg.L / mol. K

V = nRT / P

V = (1.9 x 62.363 x 294) / 697

V = 49.98 L

New pressure of the gas, P = 795 mmHg

New temperature of the gas, T = 26 °C = 273 + 26 = 299 K

New moles of the gas, n = ?

Volume of the gas is constant because volume of the flask is the same when more gas was added.

n = PV / RT

n = (795 x 49.98) / (62.363 x 299)

n = 2.13 moles

Therefore, the new moles of the gas in the flask is 2.13 moles.

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Beryllium Oxide has a molecular formula Be-O

% covalent character in Be-O = 100 - % ionic character in Be-O

Now,

% ionic character = {1 - exp[(-0.25)(E(Be)-E(O))²]}*100

where,

E(Be) and E(O) are the electronegativity values of Be and O

Based on the Paulings scale:-

E(Be) = 1.57

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% ionic = {1 - exp[(-0.25)(1.57-3.44)²]}*100 = 58.28%

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234 g

Explanation:

We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.

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1. Calculate the <em>moles of Na </em>

Moles Na = 92.0 × 1/22.99

Moles Na = 4.002 mol Na

===============

2. Calculate the <em>moles of NaCl </em>

The molar ratio is 2 mol NaCl:2 mol Na.

Moles of NaCl = 4.002 × 2/2

Moles of NaCl = 4.002 mol NaCl

===============

3. Calculate the <em>mass of NaCl </em>

Mass of NaCl = 4.002 × 58.44

Mass of NaCl = 234 g

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