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4 years ago

PLEASE HELP ME ITS EASY MY BRAIN STOPED WORKINGG :(

Chemistry

1 answer:

Zinaida [17]4 years ago

5 0

Answer:

See below!

Explanation:

Solids have a definite mass, a definite shape, and a definite volume. They are not compressible and are the most tightly packed together.

Liquids have definite mass and definite volume. Liquids take up the shape of the container and are not compressible. They are slightly packed together but are allowed to move freely.

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Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial conc

Anni [7]

Answer:

Ka = 4.76108

Explanation:

CO(g) + 2H2(g) ↔ CH3OH(g)

∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]

[ ]initial change [ ]eq

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3 years ago

Assume that 50.0mL 50.0mL of 1.0MNaCl(aq) 1.0MNaCl(aq) and 50.0mL 50.0mL of 1.0M AgNO 3 (aq) 1.0MAgNO3(aq) were combined. Accord

S_A_V [24]

Answer:

The amount of precipitate formed would 7.175 grams of silver chloride.

Explanation:

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Volume of silver nitrate solution = 50.0 mL = 0.050 L

Molarity of the silver nitrate = 1.0 M

$n'=1.0 M\times 0.050 L=0.050 mol$

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According to reaction, 1 mole of of silver nitrate reacts with 1 mole of NaCl. Then 0.050 mole of silve nitrate will :

$\frac{1}{1}\times 0.050 mol=0.050 mol$ of NaCl

This means that silver nitrate is in limiting amount and NaCl is in excessive amount.

So, the amount of AgCl depends upon amount of silver nitrate.

According to reaction, 1 mole of silver nitrate gives 1 mole of AgCl.

Then 0.050 moles of silver nitrate will give;

$\frac{1}{1}\times 0.050 mol=0.050 mol$ of AgCl

Mass of 0.050 moles of AgCl ;

$0.050 mol\times 143.5 g/mol=7.175 g$

The amount of precipitate formed would 7.175 grams of silver chloride.

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