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3 years ago

How many grams of carbon are present in 45.0 g of CCl4?

1 answer:

lisabon 2012 [21]3 years ago

5 0

To determine the amount of a certain element in a compound, we use the ratio of the elements from the compound. We calculate is follows:

45.0 g CCl4 ( 1 mol CCl4 / 153.82 g CCl4 ) ( 1 mol C / 1 mol CCl4 ) ( 12.01 g C / 1 mol C ) = 3.5135 g carbon present

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Name some acids and in what and how we used them

Ne4ueva [31]

Answer:

Acid Uses

Organic acids

Citric acid As a preservative for food As a flavouring agent

Ascorbic acid (also called vitamin C) In the treatment of bone marrow and scurvy diseases

Acetic acid Added to pickles to make them sour

Explanation:

7 0

3 years ago

Select the correct answer. Given: 2Al + 6HCl → 2AlCl3 + 3H2 If the chemical reaction produces 129 grams of AlCl3, how many grams

ivann1987 [24]

The reaction produces 2.93 g H₂.

M_r: 133.34 2.016

2Al + 6HCl → 2AlCl₃ + 3H₂

Moles of AlCl₃ = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃

Moles of H₂ = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂

Mass of H₂ = 1.451 mol H₂ × (2.016 g H₂/1 mol H₂) = 2.93 g H₂

7 0

4 years ago

What happens if you cross a heterozygous organism with a heterozygous organism?

maw [93]

If you cross a heterozygous organism with another heterozygous organism, pair two Tt and Tt. Below is a punnet square to show the result:

T t
T TT Tt
t Tt tt

This means that you have two heterozygous recessive and one homogenous dominant and one homogenous recessive.

5 0

3 years ago

5.2 X 1026 molecules of CH4

olya-2409 [2.1K]

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I cant send this with no text 

Have a nice day

8 0

3 years ago

The specific heat of copper is 0.093 cal/g0C. Calculate the temperature change that occurs if 28 g of copper at 25 0C absorbs 58

Umnica [9.8K]

Answer:

22.27 °C = ΔT

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m × c × ΔT

Given data:

mass = 28 g

heat absorbed = 58 cal

specific heat of copper = 0.093 cal/g .°C

temperature change =ΔT= ?

Solution:

Q = m × c × ΔT

58 cal = 28 g × 0.093 cal /g.°C × ΔT

58 cal = 2.604 cal.°C × ΔT

58 cal / 2.604 cal .°C = ΔT

22.27 °C = ΔT

5 0

3 years ago