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4 years ago

. How many moles of ammonia gas, NH3, are required to fill a volume of 50 liters at STP?

Chemistry

2 answers:

murzikaleks [220]4 years ago

6 0

In STP every 22.4 litters is 1 mol

KATRIN_1 [288]4 years ago

5 0

Answer : The number of moles of ammonia gas is, 2.232 moles

Explanation : Given,

Volume of ammonia gas = 50 L

As we know that at STP,

1 mole of gas contains 22.4 L volume of gas.

As, 22.4 L volume of ammonia gas present in 1 mole of ammonia gas

So, 50 L volume of ammonia gas present in $\frac{50}{22.4}=2.232$ mole of ammonia gas

Therefore, the number of moles of ammonia gas is, 2.232 moles

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3 years ago

I am having some trouble figuring out how to approach the following problem: A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is ti

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3 years ago

A technician needs 500.0mL of a 0.500 M MgCl2 solution for some lab procedures. The stock bottle is labeled 4.00 M MgCl2. What v

mihalych1998 [28]

Answer:

Explanation:

To solve this problem, we need to obtain the number of moles of the solute we desired to prepare;

Number of moles = molarity x volume

Parameters given;

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4 years ago

Convert 1.248×1010 g to each of the following units.

Leona [35]

Answer:

a) 1.248 x 10⁷ kg

b) 1.248 x 10⁴ Mg

c) 1.248 x 10¹³ mg

d) 1.248 x 10⁴ ton

Explanation:

a) Since 1000 g = 1 kg we can convert grams to kg by multiplyig any given quantity in grams by the conversion factor ( 1 kg / 1000 g):

1.248 x 10¹⁰ g * (1 kg / 1000 g) = 1.248 x 10⁷ kg

b) Since 1 Mg = 1 x 10⁶ g, the conversion factor will be ( 1 Mg / 1 x 10⁶ g):

1.248 x 10¹⁰ g * ( 1 Mg / 1 x 10⁶ g) = 1.248 x 10⁴ Mg

c) Since 1 mg = 1 x 10⁻³ g, the conversion factor will be ( 1 mg / 1 x 10⁻³ g):

1.248 x 10¹⁰ g ( 1 mg / 1 x 10⁻³ g) = 1.248 x 10¹³ mg

d) Since 1 metric ton = 1000 kg and 1000 g = 1 kg, we can use these conversions factors: ( 1 kg / 1000 g) and (1 ton / 1000 kg):

1.248 x 10¹⁰ g * ( 1 kg / 1000 g) * ( 1 ton / 1000 kg) = 1.248 x 10⁴ ton

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3 years ago