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Vilka [71]
3 years ago
11

The metals in Groups 1A, 2A, and 3A ____.

Chemistry
1 answer:
Ket [755]3 years ago
6 0
The correct answer is D.


I hope that helped! c:
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3.785412e+7cm³

Explanation:

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An unknown compound with a molar mass of 223.94 g/mol consists of 32.18% c, 4.50% h, and 63.32% cl. find the molecular formula f
Dima020 [189]

The actual number of atoms of each element present in the molecule of the compound is represented by the formula known as molecular formula.

Molar mass of the unknown compound = 223.94 g/mol (given)

Mass of each element present in the unknown compound is determined as:

  • Mass of carbon, C:

\frac{32.18}{100}\times 223.94 = 72.06 g

  • Mass of hydrogen, H:

\frac{4.5}{100}\times 223.94 = 10.08 g

  • Mass of chlorine, Cl:

\frac{63.32}{100}\times 223.94 = 141.79 g

Now, the number of each element in the unknown compound is determined by the formula:

number of moles = \frac{given mass}{molar mass}

  • Number of moles of C:

number of moles = \frac{72.06}{12} = 6.005 mole\simeq 6 mole

  • Number of moles of H:

number of moles = \frac{10.08}{1} = 10.08 mole\simeq 10 mole

  • Number of moles of Cl

number of moles = \frac{141.79}{35.5} = 3.99 mole\simeq 4 mole

Dividing each mole with the smallest number of mole, to determine the empirical formula:

C_{\frac{6}{4}}H_{\frac{10}{4}}Cl_{\frac{10}{4}}

C_{1.5}H_{2.5}Cl_{1}

Multiplying with 2 to convert the numbers in formula into a whole number:

So, the empirical formula is C_{3}H_{5}Cl_{2}.

Empirical mass = 12\times 3+1\times 5+2\times 35.5 = 112 g/mol

In order to determine the molecular formula:

n = \frac{molar mass}{empirical mass}

n = \frac{223.94}{112} = 1.99 \simeq 2

So, the molecular formula is:

2\times C_{3}H_{5}Cl_{2} =  C_{6}H_{10}Cl_{4}

6 0
3 years ago
A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita
mestny [16]

Answer:

The answer is:

(a) NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of AgNO_3,

= 30.0 ml

or,

= 0.030 L

Molarity of AgNO_3,

= 0.050 M

Moles of NaCl will be:

= \frac{Given \ mass}{Molar \ mass}

= \frac{0.0860}{58.44}

= 0.00147 \ mol

now,

Moles of AgNO_3 will be:

= Molarity\times Volume

= 0.050\times 0.030

=0.0015 \ mol

(a)

The reaction is:

⇒ NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b)

1 mole of NaCl react with,

= 1 mol of AgNO_3

0.0015 mol AgNO_3 needs,

= 0.00150 \ mol \ NaCl

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

= 0.00147\times \frac{1}{1}\times \frac{143.32}{1}

= 0.211 \ g \ AgCl

4 0
3 years ago
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