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3 years ago

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All salts are ionic compounds, but not all ionic compounds are salts. All salts are ionic compounds, but not all ionic compounds

are salts.
a. True
b. False

1 answer:

allsm [11]3 years ago

3 0

I think this question is a true statement :)

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) Consider the starting materials and reagents. What do you expect to happen at the beginning of this reaction (Boxes 1-3 on the

tekilochka [14]

Answer:

Check the explanation

Explanation:

functional group found in the major organic product = alpha -beta unsaturated ketone

Reaction used to form this functional group = Michael condensation reaction

Also other reactions are - Aldol condensation , Robinson annulation reaction.

Kindly check the attached image below to see the step by step solution to the question above.

3 0

3 years ago

A 0.1014 g sample of a purified compound containing C, H, and, O was burned in a combustion apparatus and produced 0.1486 g CO2

Alina [70]

Answer: The empirical formula for the given compound is $CH_2O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=0.1486g$

Mass of $H_2O=0.0609g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 0.1486 g of carbon dioxide, $\frac{12}{44}\times 0.1486=0.0405g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.0609 g of water, $\frac{2}{18}\times 0.0609=0.00677$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1014) - (0.0405 + 0.00677) = 0.054 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.0405g}{12g/mole}=0.003375moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.00677g}{1g/mole}=0.00677moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.054g}{16g/mole}=0.003375moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.485 moles.

For Carbon = $\frac{0.003375}{0.003375}=1$

For Hydrogen = $\frac{0.00677}{0.003375}=2.00\times 2$

For Oxygen = $\frac{0.003375}{0.003375}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 2 : 1

The empirical formula for the given compound is $C_1H_2O_1=CH_2O$

Thus, the empirical formula for the given compound is $CH_2O$

8 0

3 years ago

If you start a chemical reaction with 5 atoms of helium, how many atoms will you have after?

Ivenika [448]

Answer:

5 atoms

Explanation:

When a chemical reaction starts with 5 atoms of hydrogen, and the reaction goes into completion, it must give us 5 atoms of hydrogen.

This is in compliance with the law of conservation of mass which states that "in a chemical reaction, matter is neither created nor destroyed but converted from one form to another".

In chemical reactions, bonds are broken and new compounds are formed in the process. The same number of species at the start and end of the reaction must still be the same.

3 0

2 years ago

Read 2 more answers

3. What is an isotope?

prohojiy [21]

Answer:

one of two or more species of atoms of a chemical element with the same atomic number and position in the periodic table and nearly identical chemical behaviour but with different atomic masses and physical properties

Explanation:

6 0

3 years ago

Read 2 more answers

0.556 g of a solid white acid are dissolved in water and completely neutralized by the addition of 52.38 mL of 0.396 M NaOH. Cal

Veseljchak [2.6K]

Answer:

$M_{acid}=26.804g/mol$

Explanation:

Hello,

In this case, by knowing that the used NaOH equals in moles the acid (monoprotic) as shown below, during the titration:

$n_{acid}=n_{NaOH}$

By knowing the volume and the concentration of the NaOH, one obtains:

$n_{acid}=0.05238L*0.396\frac{mol}{L}=0.0207mol$

Thus, the molar mass of the acid is computed based on the previously computed moles and the given mass as follows:

$M_{acid}=\frac{m_{acid}}{n_{acid}}=\frac{0.556g}{0.0207mol}=26.804g/mol$

Best regards.

7 0

3 years ago