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2 years ago

Atoms combine to form?

1 answer:

5 0

Atoms combine to form molecules, but your answer could also be elements. I can't be sure because you gave no context. Let me know if I can elaborate in any way!

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Which statement about fission is correct?

sashaice [31]

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3 years ago

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Match the name of the greenhouse gas with one of the human activities that produces it.

ss7ja [257]

Answer:

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3 years ago

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. What is a carbonyl group? Draw the carbonyl groups that are characteristic of aldehydes, ketones, carboxylic acids, and esters

Elan Coil [88]

Answer :

Carbonyl group : It is a functional group composed of a carbon atom that double bonded to oxygen atom. It is represented as $C=O$

Carboxylic group : It is the class of organic compound in which the carboxylic (-COOH) group is attached to a hydrocarbon is known as carboxylic.

The general formula of carboxylic is, $C_nH_{2n+1}COOH$ . According to the IUPAC naming, the carboxylic are named as alkanoic acids.

Aldehyde group : It is the class of organic compound in which the (-CHO) group is attached to a hydrocarbon is known as aldehyde.

The general representation of aldehyde is, $R-CHO$ . According to the IUPAC naming, the aldehyde are named as alkanals.

Ketone group : It is the class of organic compound in which the (-CO) group is directly attached to the two alkyl group of carbon is known as ketone.

The general representation of ketone is, $R-CO-R^'$ . According to the IUPAC naming, the ketone are named as alkanone.

Ester group : It is the class of organic compound in which the (-COO) group is directly attached to the two alkyl group of carbon is known as ester.

The general representation of ester is, $R-COO-R^'$ . According to the IUPAC naming, the ester are named as alkyl alkanoate.

6 0

3 years ago

How many milligrams of MgI2 must be added to 257.7 mL of 0.087 M KI to produce a solution with [I−] = 0.1000 M?

lbvjy [14]

Answer:

The answer is 465.6 mg of MgI₂ to be added.

Explanation:

We find the mole of ion I⁻ in the final solution

C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol

But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.

So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.

Hence, the weight of MgI₂ must be added is

Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg

4 0

2 years ago

combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon

masya89 [10]

Answer:

$\rm CH_2$ .

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be $\rm CO_2$ and $\rm H_2O$ . The $\rm C$ and $\rm H$ would come from the hydrocarbon, while the $\rm O$ atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

$\rm C$ : $12.011$ .
$\rm H$ : $1.008$ .
$\rm O$ : $\rm 15.999$ .

Calculate the molar mass of $\rm CO_2$ and $\rm H_2O$ :

$M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$

Calculate the number of moles of $\rm CO_2$ molecules in $33.01\; \rm g$ of $\rm CO_2\!$ :

$\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol$ .

Similarly, calculate the number of moles of $\rm H_2O$ molecules in $13.51\; \rm g$ of $\rm H_2O\!$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol$ .

Note that there is one carbon atom in every $\rm CO_2$ molecule. Approximately $0.7501\; \rm mol$ of $\rm CO_2\!$ molecules would correspond to the same number of $\rm C$ atoms. That is: $n(\mathrm{C}) \approx 0.7501\; \rm mol$ .

On the other hand, there are two hydrogen atoms in every $\rm H_2O$ molecule. approximately $0.7499\; \rm mol$ of $\rm H_2O$ molecules would correspond to twice as many $\rm H\!$ atoms. That is: $n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol$ .

The ratio between the two is: $n(\mathrm{C}): n(\mathrm{H}) \approx 1:2$ .

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be $\rm CH_2$ .

6 0

3 years ago