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Hatshy [7]
3 years ago
15

A container of oxygen gas is at STP. If this sample is put into an oven at 280 C, what would its pressure be, in atmospheres?

Chemistry
2 answers:
Leni [432]3 years ago
3 0

Explanation:

Step 1:

Data obtained from the question. This include the following:

Initial pressure (P1) = 1atm

Initial temperature (T1) = 0°C = 0°C + 273 = 273K

Final temperature (T2) = 280°C = 280°C + 273 = 553K

Final pressure (P2) =...?

Step 2:

Determination of the new pressure of the gas.

Since the volume of the gas is constant, the following equation:

P1/T1 = P2/T2

will be used to obtain the pressure. This is illustrated below:

P1/T1 = P2/T2

1/273 = P2 / 553

Cross multiply

273x P2 = 553

Divide both side by 273

P2 = 553/273

P2 = 2.03atm

Therefore, the new pressure of the gas will be 2.03atm

saw5 [17]3 years ago
3 0

Answer:

\boxed {\boxed {\sf P_2=2.03 \ atm}}

Explanation:

We are concerned with the variables of temperature and pressure, so we use Gay-Lussac's Law, which states the temperature of a gas is directly proportional to the pressure. The formula is:

\frac{P_1}{T_1}=\frac{P_2}{T_2}

We know that the container of gas begins at standard temperature and pressure (STP). This is 1 atmosphere of pressure and 273 Kelvin.

\frac { 1 \ atm}{ 273 \ K} = \frac{P_2}{T_2}

We know the gas is put into an oven at 280 degrees Celsius. We can convert this to Kelvin.

  • K= °C + 273.15
  • K= 280 +273.15
  • K= 553.15

\frac { 1 \ atm}{ 273 \ K} = \frac{P_2}{553.15 \ K}

We are solving for the new pressure, so we must isolate the variable P₂. It is being divided by 553.15 Kelvin. The inverse of division is multiplication, so we multiply both sides by 553.15 K

553.15 \ K *\frac { 1 \ atm}{ 273 \ K} = \frac{P_2}{553.15 \ K} * 553.15 \ K

553.15 \ K *\frac { 1 \ atm}{ 273 \ K}= P_2

The units of Kelvin cancel.

553.15  *\frac { 1 \ atm}{ 273 }= P_2

2.02619047619 \ atm = P_2

Rounded to the nearest hundredth:

2.03   \ atm \approx P_2

The new pressure is approximately <u>2.03 atmospheres.</u>

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8 0
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Consider the following reaction. Zn(s) 2agno3(aq) --&gt; 2ag(s) zn(no3)2(aq)when 16. 2 g of silver was produced, _____ mole(s) o
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The required amount of silver nitrate to produce 16.2g of silver is 25.48 grams.

<h3>What is the relation between mass & moles?</h3>

Relation between the mass and moles of any substance will be represented as:

n = W/M, where

  • W = given mass
  • M = molar mass

Moles of silver = 16.2g / 107.8g/mol = 0.15mol

From the stoichiometry of the given reaction it is clear that, same moles of silver nitrate is required to produce same moles of silver. So 0.15 moles of silver nitrate is required.

Mass of silver nitrate = (0.15mol)(169.87g/mol) = 25.48g

Hence required mass of silver nitrate is 25.48g.

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brainly.com/question/19784089

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5 0
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Gastric juice in the stomach contains pepsin and sulfuric acid.<br><br><br> TrueFalse
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5 0
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When silver nitrate reacts with copper, copper(II) nitrate and silver are produced. The balanced equation for this reaction is:
posledela

The question is incomplete, here is the complete question:

When silver nitrate reacts with copper, copper(II) nitrate and silver are produced. The balanced equation for this reaction is:

2AgNO_3+Cu\rightarrow Cu(NO_3)_2+2Ag

Suppose 6 moles of silver nitrate react. The reaction consumes___ moles of copper. The reaction produces __ moles of copper(II) nitrate and __ moles of silver.

<u>Answer:</u> The amount of copper metal reacted is 3 moles, amount of copper (II) nitrate produced is 3 moles and amount of silver metal produced is 6 moles.

<u>Explanation:</u>

We are given:

Moles of silver nitrate = 6 moles

For the given chemical reaction:

2AgNO_3+Cu\rightarrow Cu(NO_3)_2+2Ag

  • <u>For copper metal:</u>

By Stoichiometry of the reaction:

2 moles of silver nitrate reacts with 1 mole of copper metal

So, 6 moles of silver nitrate will react with = \frac{1}{2}\times 6=3mol of copper metal

Moles of copper reacted = 3 moles

  • <u>For copper(II) nitrate:</u>

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 1 mole of copper(II) nitrate

So, 6 moles of silver nitrate will produce = \frac{1}{2}\times 6=3mol of copper(II) nitrate

Moles of copper(II) nitrate produced = 3 moles

  • <u>For silver metal:</u>

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver metal

So, 6 moles of silver nitrate will produce = \frac{2}{2}\times 6=6mol of silver metal

Moles of silver metal produced = 3 moles

Hence, the amount of copper metal reacted is 3 moles, amount of copper (II) nitrate produced is 3 moles and amount of silver metal produced is 6 moles.

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Answer:

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5 0
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