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Radda [10]
2 years ago
10

The elemental composition of a compound is determined to be 39.5% oxygen, 32.1% chromium and 28.4% sodium by weight.

Chemistry
1 answer:
maksim [4K]2 years ago
3 0

Answer:

Explanation:

To calculate percent composition, we divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:

​%C=

​12.04g compound

​

​7.34g C

​​ ×100%=61.0%

​%H=

​12.04g compound

​

​1.85g H

​​ ×100%=15.4%

​%N=

​12.04g compound

​

​2.85g N

​​ ×100%=23.7%

​​  

The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.

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The answer is protons. Neutrons have no charge and electrons have a negative charge so the positive charge must be protons.
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A compound with the empirical formula ch2 has a molar mass of 70 g/mol. what is the molecular formula for this compound?
Rainbow [258]

The empirical formula, <span>C<span>H2</span></span>, has a relative molecular mass of

<span>1×<span>(12.01)</span>+2×<span>(1.01)</span>=14.04</span>

This means that the empirical formula must be multiplied by a factor to bring up its molecular weight to 70. This factor can be calculated as the ratio of the relative masses of the molecular and empirical formulas

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Remember that subscripts in molecular formulas must be in whole numbers, hence the rounding-off. Finally, the molecular formula is

<span><span>C<span>1×5</span></span><span>H<span>2×5</span></span>=<span>C5</span><span>H<span>10</span></span></span>

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3 years ago
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26.5 g of a solution with a density of 7.48 g/mL
Cloud [144]

Answer:  Assuming the question: 3.54 ml (3 sig figs)

Explanation:

I don't see a question, but will assume it is "What volume is needed to obtain 26.5 grams?

If so:

(26.5 g)/(7.48 ml) = 3.54 ml (3 sig figs)

6 0
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A student plans to use a density versus solution concentration standard curve to identify the sodium chloride concentration in a
Art [367]

Answer:

a. 0.50 g, 1.0 g, 1.5 g, 2.0 g, 2.5 g;

g. 0.1 g, 0.5 g, 1.0 g, 1.5 g, 2.0 g

Explanation:

The percent mass is defined as a ratio between the mass of a solute and mass of a solution:

\omega = \frac{m_{solute}}{m_{solution}}\cdot 100\%

Since solution only consists of a solute and solvent, express its mass as:

m_{solution} = m_{solute} + m_{solvent}

Then:

\omega = \frac{m_{solute}}{m_{solution}}\cdot 100\%=\frac{m_{solute}}{m_{solute} + m_{solvent}}

Firstly, solve for how much mass is required to prepare 3.0 %. Let's say, we have x g of the solute:

0.03 = \frac{x}{30.0 + x}\therefore x = 0.03(30.0 + x)

x = 0.90 + 0.03x

0.97x = 0.90\therefore x = 0.93 g

Similarly, solve for 6.0 %, let's say, we have x g of the solute again:

0.06 = \frac{x}{30.0 + x}\therefore x = 0.06(30.0 + x)

x = 1.80 + 0.06x

0.94x = 1.80\therefore x = 1.91 g

Hence, masses should be in a range of 0.93 g to 1.91 g.

7 0
3 years ago
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