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adelina 88 [10]
2 years ago
14

Calcium carbide, CaC2, reacts with water to form calcium hydroxide and the flammable gas ethyne (acetylene) in the reaction: Wha

t mass of ethyne can be produced
Chemistry
1 answer:
AysviL [449]2 years ago
7 0

Answer:

1 mole of CaC₂ will produce 26g of C₂H₂ or 64.1g of CaC₂ will produce 26g of C₂H₂

Explanation:

Hello,

To solve this question, we'll require a balanced chemical equation of reaction between calcium carbide and water.

Equation of reaction

CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂

Molar mass of calcium carbide (CaC₂) = 64.1g/mol

Molar mass of water (H₂O) = 18g/mol

Molar mass of calcium hydroxide (Ca(OH)₂) = 74g/mol

Molar mass of ethyne (C₂H₂) = 26g/mol

From the equation of reaction, 1 mole of CaC₂ will produce 1 mole of C₂H₂

1 mole of CaC₂ = mass / molar mass

Mass = 1 × 64.1

Mass = 64.1g

1 mole of C₂H₂ = mass / molar mass

Mass = 1 × 26

Mass = 26g

Therefore, 1 mole of CaC₂ will produce 26g of C₂H₂

Note: this is a hypothetical calculation since we were not given the initial mass of CaC₂ that starts the reaction

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AleksandrR [38]

The element should be placed in the column for metalloids. Option A

<h3>What is the property of metals?</h3>

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Metalloids are generally amphoteric in nature. They can react with both the acid and the basses to form compounds. The nonmetals do not react with acid or base.

Looking at the result, we are told that the entry in the second row have the entries  nothing, nothing, nothing. If this is the outcome, then it follows that the element should be placed in the column for metalloids.

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4 0
1 year ago
Mass x height x gravity is the formula to calculate:
Sindrei [870]

Answer:

Potential energy

Explanation:

p.e = mgh

8 0
2 years ago
Which of the following displays the correct change in enthalpy and best describes the reaction below? 2CsCl(aq) + Na2SO4(aq) 2Na
Bond [772]
We subtract the enthalpies of the reactants from that of the products:
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3 0
3 years ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
The concentration of a solution can be expressed in
puteri [66]
The concentration of a solution can be expressed in (4) <span>moles per liter~</span>
3 0
3 years ago
Read 2 more answers
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