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3 years ago

What happens during a chemical change??

Chemistry

2 answers:

Alenkasestr [34]3 years ago

6 0

I think it's b because of the different choices you have don't make sense to me

borishaifa [10]3 years ago

5 0

Hmmm.. i think B. I dont understand the the answer choices.

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What is the noble gas configuration of lithium? (li)?

Nat2105 [25]

The noble gas configuration means that the electronic configuration for Lithium is written in shortcut by starting with the symbol for the nearest noble gas. Note that the noble gas should be lighter than the said element. In here, the nearest noble gas to Lithium is Helium (He). Then, add the additional electrons. Thus, the noble gas configuration would be: <em>[He]2s¹.</em>

8 0

3 years ago

Determine the enthalpy change for the decomposition of calcium carbonate. CaCO₃(s) --> CaO(s) + CO₂(g) given the thermochemic

Hunter-Best [27]

Answer: c. 179 kJ/mol

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to Hess’s law, the chemical equation can be treated as algebraic expressions and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

Given:

$Ca(OH)_2(s)\rightarrow CaO(s)+H_2O (l)$ $\Delta H_1= 65.2 kJ/mol$ (1)

$Ca(OH)_2(s)+CO_2(g)\rightarrow CaCO_3(s)+H_2O(l)$ $\Delta H_2= -113.8 kJ/mol$ (2)

$C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_3= -393.5 kJ/mol$ (3)

$2Ca(s)+O_2(g)\rightarrow 2CaO(s)$ $\Delta H_4=-1270.2 kJ/mol$ (4)

On subtracting eq (1) from eq (2) we have:

$Ca(OH)_2(s)\rightarrow CaO(s)+H_2O(l)$ - $Ca(OH)_2(s)+CO_2(g)\rightarrow CaCO_3(s)+H_2O(l)$

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

$\Delta H=-113.8-(65.2)kJ/mol=-179kJ/mol$

Hence the enthalpy change for the raection is 179.0 kJ/mol.

6 0

3 years ago

Which piece of equipment would be used to measure the volume of liquid

cestrela7 [59]

It depends on your answers. Common liquid measuring devices include Graduated Cylinders, Test Tubes, and Beakers. The most likely answer to your question would be a Graduated Cylinder.

3 0

3 years ago

Select three parts of the cell theory.

Sergio [31]

The answer is
A,B,C

all cells come from cells.

All living things are made up of cells.

A cell is the smallest unit of life.

8 0

3 years ago

A 35.161 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion analysis ap

xxMikexx [17]

Answer:

The empirical formulae is C6H12S02

Explanation:

1. First we need to obtain the mass of each element in the sample and compound formed

Carbon = (62.637 mg * 12.011 g/mol / 44.009 g/mol) = 17.094 mg of Carbon

Hydrogen = ( 25.641 mg * (2 *1..008 g/mol) / 18.015 g/mol) = 2.869 mg of Hydrogen

Sulphur = (13.54 mg * 32.066 g/mol / 64.066 g/mol) = 6.777 mg of Sulphur

2. Next is to determine the percentage composition. Here we divide the respective mass by the mass of the sample

Carbon = 17.094 / 35.161 * 100 = 48.62 %

Hydrogen = 2.869/ 35.161 *100 = 8.16 %

Sulphur = 6.777/ 31.321 *100 = 21.64 %

Oxygen = (100 - (48.62 + 8.16 + 21.64)) = 21.58 %

3. Next is to divide the mass assuming there are 100 mg by the respective atomic masses to obtain the number of moles

Carbon = 48.62 / 12.011 = 4.048 mol

Hydrogen = 8.16 / 1.008 = 8.095 mol

Sulphur = 21.64 / 32.066 = 0.675 mol

Oxygen = 21.58 / 16.000 = 1.348 mol

Next is to divide by the smallest value

Carbon = 4.048/ 0.675 =5.997 = 6

Hydrogen = 8.095 / 0.675 =11.993 =12

Sulphur = 0.675/ 0.675 = 1

Oxygen = 1.348 / 0.675 = 1.997 = 2

So therefore the empirical formulae of the sample is C6H12SO2

7 0

4 years ago