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tatiyna
3 years ago
11

Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing

carbon dioxide gas and water vapor. Calculate the moles of acetylene needed to produce of carbon dioxide. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.
Chemistry
1 answer:
serg [7]3 years ago
5 0

Answer:

2.25 moles ≈ 2.3 moles

Explanation:

Full Question:

Acetylene C2H2 gas is often used in welding torches because of the very high heat produced when it reacts with oxygen O2 gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce 1.5mol of water. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.

First off, let's put the equation of the reaction sown;

C2H2 + O2 → H2O + CO2

Upon balancing, we have;

2C2H2 + 3O2 → 2H2O + 4CO2

So from the equation;

3 mole of oxygen is required in order to produce 2 mole of water vapour.

The question asks to calculate moles of oxygen needed to produce 1.5mol of water

This leads us to;

3 moles = 2 moles

x moles = 1.5 moles

x = ( 1.5 * 3 ) / 2

x = 4.5 / 2 = 2.25 moles ≈ 2.3 moles

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Answer:

The mol fraction of sucrose in an aqueous solution at 20°C  is 0.114

Explanation:

Step 1: Data given

The vapor pressure of water above the solution is 2.0 mmHg

The vapor pressure of pure water at 20 °C is 17.5 mmHg

Step 2: Calculate the mol fraction of sucrose

Psolution = (χsolvent) (P°solvent)

⇒with Psolution = the vapor pressure of the solution is 2.0 mmHg

⇒with χsolvent = the mol fraction of sucrose = TO BE DETERMINED

⇒with P°solvent = the vapor pressure of pure water at 20 °C =17.5 mmHg

2.0 mmHg = X * 17.5 mmHg

X = 2.0 mmHg / 17.5 mmHg

X = 0.114

The mol fraction of sucrose in an aqueous solution at 20°C  is 0.114

7 0
3 years ago
What is the mass of a sample of NH3 containing 7.20 x 1024 molecules of NH3? (5 points) Group of answer choices 203 grams 161 gr
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Answer:

203 grams

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The no. of grams of NH₃ present = no. of moles x molar mass = (11.96 mol)(17.0 g/mol) = 203.3 g ≅ 203.0 g.

8 0
3 years ago
Identify oxidation.
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Oxidation is "Increase in oxidation number" as well as loss of electrons.

A rise in oxidation number results from the loss of negative electrons, whereas a reduction in oxidation number results from the gain of electrons. As a consequence, the oxidized element or ion experiences a rise in oxidation number.

As a result of losing electrons in the process, a reactant oxidizes. When a reactant obtains electrons during a reaction, reduction takes place. This frequently happens when acid and metals react.

Therefore, Oxidation is "Increase in oxidation number" as well as loss of electrons.

Hence, the correct answer will be option (e)

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brainly.com/question/16976470

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Answer:

Your statement is right.

Explanation:

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