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3 years ago

_____ are a type of organic compound.

2 answers:

8 0

Hydrocarbons are a type of organic compound. The answer is letter C. Hydrocarbons contains a carbon, hydrogen and oxygen bond that makes it organic. Chlorides, carbonates and nitrous oxides are not organic compounds but they are inorganic compounds.

nikdorinn [45]3 years ago

3 0

<span>A hydrocarbon is a molecule whose structure includes only hydrogen and carbon atoms. Hydrocarbons containing between six and ten carbon molecules are the top components of most fuels. In general, these molecules are burned to produce energy. Examples such include: 1. Methane: A fuel in electrical generation. 2. Propane: Generally used heating and cooking. 3. Butane: Generally used in lighters and aerosol cans. 4.Heptane: The major component of gasoline Hydrocarbons are majorly used as a combustible fuel source and are considered a primary source of energy for contemporary civilizations. Oil is an excellent example of a hydrocarbon fuel currently in use all over the world.</span>

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1 2 3 4 5 6 7 8 9 10TIME REMAINING53:56A scientist told his student assistants that he had sorted groups of mice by their intell

Gekata [30.6K]

Answer:

People can introduce their own biases into an experiment.

Explanation:

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2 years ago

What would be oil classification?

boyakko [2]

The service rating of passengers car and commercial automotive motor oils

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3 years ago

Correction?

rjkz [21]

Your answer is correct.

MgO(s) + H2O(l) ----> Mg(OH)2(aq)

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3 years ago

Read 2 more answers

Substance X is a compound containing 632mg of manganese and 368mg of oxygen. Substance X is shown

defon

The empirical formula : MnO₂.

<h3>Further explanation</h3>

Given

632mg of manganese(Mn) = 0.632 g

368mg of oxygen(O) = 0.368 g

M Mn = 55

M O = 16

Required

The empirical formula

Solution

You didn't include the pictures, but the steps for finding the empirical formula are generally the same

Find mol(mass : atomic mass)

Mn : 0.632 : 55 = 0.0115

O : 0.368 : 16 =0.023

Divide by the smallest mol(Mn=0.0115)

Mn : O =

$\tt \dfrac{0.0115}{0.0115}\div \dfrac{0.023}{0.0115}=1\div 2$

The empirical formula : MnO₂

8 0

3 years ago

What is the percent yield of LiCl if I produced 30.85g LiCl and my theoretical yield was calculated to be 35.40g LiCl?

marissa [1.9K]

Answer:

87.15%

Explanation:

To find percent yield, we can use this simple equation

$\frac{Actual}{Theoretical} *100$

Where "Actual" is the amount in grams actually collected from the reaction, and "Theoretical" is, well, the theoretical amount that should have been produced.

They give us these values, so to find the percent yield, just plug the numbers in.

$\frac{30.85}{35.40} *100\\\\ =87.15$

So, the percent yield is 87.15%

An easy trick to remember how to do this is just to divide the smaller number by the bigger number and move the decimal back two places. If you have a percent yield greater than 100%, something is wrong in the reaction.

7 0

3 years ago