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3 years ago

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Part APart complete A sample of sodium reacts completely with 0.568 kg of chlorine, forming 936 g of sodium chloride. What mass

of sodium reacted? Express your answer to three significant figures and include the appropriate units.

1 answer:

yan [13]3 years ago

4 0

Answer: 368 grams of sodium reacted.

Explanation:

The balanced reaction is :

$2Na(s)+Cl_2(g)\rightarrow 2NaCl(s)$

$\text{Moles of chlorine}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of chlorine}=\frac{0.568\times 1000g}{71g/mol}=8moles$

$\text{Moles of sodium chloride}=\frac{936g}{58.5g/mol}=16mol$

According to stoichiometry :

2 moles of $NaCl$ are formed from = 2 moles of $Na$

Thus 16 moles of $NaCl$ are formed from= $\frac{2}{2}\times 16=16moles$ of $Na$

Mass of $Na=moles\times {\text {Molar mass}}=16moles\times 23g/mol=368g$

Thus 368 grams of sodium reacted.

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Group 1 elements have an average electronegativity of 0.84 (not including hydrogen). Group 17 elements have an average electrone

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We have that the Electronegative difference of 2.15 basically makes ionic bonds

Ionic bonds is to be formed

<h3>Bonding</h3>

Generally, the common electronegativity of the team 1 <u>factors </u>are 0.84 and the common <em>electronegativity </em>of the team 17 factors are 2.99.

Group 17 factors are an awful lot greater electronegative than Group 1 elements.

Group 17 consists of F, Cl, Br and I,Group 1 factor that is Na loses 1 electron from its outermost orbit and varieties Na+ cation.

Group 17 Element that is Cl accepts 1 electron in its outermost orbit and Creates Cl- anion. Now, Na+ cation and Cl- anion combines collectively by using ionic bonds.

Hence, the electronegative difference

(2.99-0.84)=2.15

Electronegative difference of 2.15 basically makes ionic bonds

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3 years ago

A compound contains 1.2 g of carbon, 3.2 g of oxygen and 0.2g of hydrogen. Find the formula of the compound

Karolina [17]

Answer:

The empirical formula of the compound is $C_{0.504}HO_{1.008}$ .

Explanation:

We need to determine the empirical formula in its simplest form, where hydrogen ( $H$ ) is scaled up to a mole, since it has the molar mass, and both carbon ( $C$ ) and oxygen ( $O$ ) are also scaled up in the same magnitude. The empirical formula is of the form:

$C_{x}HO_{y}$

Where $x$ , $y$ are the number of moles of the carbon and oxygen, respectively.

The scale factor ( $r$ ), no unit, is calculated by the following formula:

$r = \frac{M_{H}}{m_{H}}$ (1)

Where:

$m_{H}$ - Mass of hydrogen, in grams.

$M_{H}$ - Molar mass of hydrogen, in grams per mole.

If we know that $M_{H} = 1.008\,\frac{g}{mol}$ and $m_{H} = 0.2\,g$ , then the scale factor is:

$r = \frac{1.008}{0.2}$

$r = 5.04$

The molar masses of carbon ( $M_{C}$ ) and oxygen ( $M_{O}$ ) are $12.011\,\frac{g}{mol}$ and $15.999\,\frac{g}{mol}$ , then, the respective numbers of moles are: ( $r = 5.04$ , $m_{C} = 1.2\,g$ , $m_{O} = 3.2\,g$ )

Carbon

$n_{C} = \frac{r\cdot m_{C}}{M_{C}}$ (2)

$n_{C} = \frac{(5.04)\cdot (1.2\,g)}{12.011\,\frac{g}{mol} }$

$n_{C} = 0.504\,moles$

Oxygen

$n_{O} = \frac{r\cdot m_{O}}{M_{O}}$ (3)

$n_{O} = \frac{(5.04)\cdot (3.2\,g)}{15.999\,\frac{g}{mol} }$

$n_{O} = 1.008\,moles$

Hence, the empirical formula of the compound is $C_{0.504}HO_{1.008}$ .

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3 years ago

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Explanation:

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Answer:

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This is the case of the scientist in this question, who is conducting an experiment to test what temperature causes the fastest growth rate in sunflower plants. He conducted the experiment multiple times, making it an example of REPETITION.

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