We need to calculate the equivalent amount in units of moles of ammonium ions from the mass units. For this we need the molar mass of the substances involved. We calculate as follows:

31.3 g (NH4)2CO3 ( 1 mol (NH4)2CO3 / 96.09 g (NH4)2CO3) ( 2 mol NH4 / 1 mol (NH4)2CO3 ) = 0.65 mol ammonium ions

7 0

3 years ago

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Write the following numbers in standard notation, maintaining the same number of significant figures.

Alexxandr [17]

Answer:

6.104x10^2=610.4

9.5x10^3=9500

Explanation:

7 0

3 years ago

Air containing 20.0 mol% water vapor at an initial pressure of 1 atm absolute is cooled in a 1- liter sealed vessel from 200°C t

chubhunter [2.5K]

Answer:

This solution is quite lengthy

Total system = nRT

n was solved to be 0.02575

nH20 = 0.2x0.02575

= 0.00515

Nair = 0.0206

PH20 = 0.19999

Pair = 1-0.19999

= 0.80001

At 15⁰c

Pair = 0.4786atm

I used antoine's equation to get pressure

The pressure = 0.50

2. Moles of water vapor = 0.0007084

Moles of condensed water = 0.0044416

Grams of condensed water = 0.07994

Please refer to attachment. All solution is in there.

6 0

3 years ago