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3 years ago

How does a decrease in temperature affect the pressure of a contained gas

1 answer:

6 0

Increases pressure (inverse relationship) because as you decrease temperature, molecules are moving slower and have less kinetic energy. Pressure decreases as you increase temperature because molecules move past each other quicker, so they don’t bump into each other with as much force.

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I really need some help and explaining with the Heat of Combustion Lab.

Nataly_w [17]

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6 0

3 years ago

Find change H for the reaction as it occurred in the calorimeter.

dimaraw [331]

Delta H = q / mass * delta temperature

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4 years ago

Day and night are caused by... 1.the tilt of Earth's axis. 2. Earth's revolution around the sun. 3. eclipses. 4.Earth's rotation

Kryger [21]

Answer:

earths rotation on its axis

Explanation:

Day and night are due to the Earth rotating on its axis, not its orbiting around the sun. The term 'one day' is determined by the time the Earth takes to rotate once on its axis and includes both day time and night time.

5 0

3 years ago

PLZ HELP ME. This is my second time posting my question because the first time a person put a random answer. If you do not know

kupik [55]

Answer:

PLZ HELP ME. This is my second time posting my question because the first time a person put a random answer. If you do not know the question please do not answer it and leave it to someone else. Thank you, and my question is on the attached image below.

5 0

3 years ago

Read 2 more answers

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.32 g of ethane is

Pie

Answer:

There will remain 8.06 grams of ethane

Explanation:

Step 1: Data given

Mass of ethane = 9.32 grams

Mass of oxygen = 12.0 grams

Molar mass ethane = 30.07 g/mol

Molar mass oxygen = 32.00 g/mol

Step 2: The balanced equation

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Step 3: Calculate moles ethane

Moles ethane = mass ethane / molar mass ethane

Moles ethane = 9.32 grams / 30.07 g/mol

Moles ethane = 0.3099 moles

Step 4: Calculate moles oxygen

Moles oxygen = 12.0 grams / 32.0 g/mol

Moles oxygen = 0.375 moles

Step 5: Calculate the limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.375 moles)

Ethane is in excess. There will react 2/7 * 0.375 = 0.107 moles

There will remain 0.375 - 0.107 = 0.268 moles

Step 6: Calculate mass ethane

Mass ethane = moles ethane * molar mass ethane

Mass ethane = 0.268 moles * 30.07 g/mol

Mass ethane = 8.06 grams

There will remain 8.06 grams of ethane

7 0

4 years ago