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Pepsi [2]
3 years ago
7

A 3.0 g sample of a gas occupies a volume of 1.00L at 100C and 740 torr pressure. The molecular weight of the

Chemistry
1 answer:
SOVA2 [1]3 years ago
3 0

Answer:

94.2 g/mol

Explanation:

Ideal Gases Law can useful to solve this

P . V = n . R . T

We need to make some conversions

740 Torr . 1 atm/ 760 Torr = 0.974 atm

100°C + 273 = 373K

Let's replace the values

0.974 atm . 1 L = n . 0.082 L.atm/ mol.K . 373K

n will determine the number of moles

(0.974 atm . 1 L) / (0.082 L.atm/ mol.K . 373K)

n = 0.032 moles

This amount is the weigh for 3 g of gas. How many grams does 1 mol weighs?

Molecular weight → g/mol → 3 g/0.032 moles = 94.2 g/mol

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Do gases have mass and weight?​
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3 years ago
Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points.
OverLord2011 [107]
The freezing point depression is a colligative property, which means that it depends on the number of particles of solute disolved in the solution.

When you have solutes that are ionic compounds they dissociate in water into ions, then the compound that dissociates more ions will produce more particles and will decrease the freezing point the most.

Given theses aqueous solutions Na2 CO3, Co Cl3, and Li NO3 you can predict the order of the freezing points.

First, write the dissociation equations>

Na2CO3 -> 2Na(+) + CO3 (2-)  These are 3 ions: two of Na(+) and one of CO3(2-)

The number inside parenthesis are number of charge not number of molecules.

Co Cl3 -> Co(3+) + 3 Cl (1-) Those are 4 ions: one of Co (+) and three of Cl (-)

Li NO3 -> Li (+) + NO3 (-) those are two ions: one of Li (+) and one of NO3(-)

Then the ionic compound that dissociates into more ions give the solution with lower freezing point, and these is the rank from higher to lower freezing point:

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7 0
4 years ago
What is the mass of 0.73 moles of AgNO3?
love history [14]

Answer:

124 g (3 sig figs)

or

124.011 g (6 sig figs

Explanation:

Step 1: Calculate g/mol for AgNO₃

Ag - 107.868 g/mol

N - 14.01 g/mol

O - 16.00 g/mol

107.868 + 14.01 + 16.00(3) = 169.878 g/mol

Step 2: Multiply 0.73 moles by molar mass

0.73 mol (169.979 g/mol)

124 grams of AgNO₃

6 0
3 years ago
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