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KIM [24]
3 years ago
12

You find a compound composed only of element X and chlorine and you know that the compound is 13.10% X by mass. Each molecule of

the compound contains six times as many chlorine atoms as X atoms. What is element X?
Chemistry
1 answer:
8_murik_8 [283]3 years ago
8 0

Answer:

Explanation:

So, the formula for the compound should be:

XCl_{6}

Now we assume that we have 1 mol of substance, so we can make calculations to know the molar mass of element X, as follows:

M_{Cl}=35.45g/mol\\

So we have that 6 moles weight 212.7g, and we can make a rule of three to know the weight of compound X:

212.7g\rightarrow 86.9\%\\x\rightarrow 13.1\%\\\\x=\frac{212.7g*13.1}{86.9} =32.06g

As we used 1 mol, we know that the molar mass is 32.06g/mol

So the element has a molar mass of 32.06 g/mol and an oxidation state of +6, with this information, we can assure that the element X is sulfur, so the compound is SCl_{6}

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artcher [175]
Hi

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6 0
3 years ago
The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 2.001 g sample of B
Alborosie

<u>Answer:</u> The empirical formula for the given compound is C_{15}H_{24}O_1

<u>Explanation:</u>

The chemical equation for the combustion of compound having carbon, hydrogen, iron 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=5.995g

Mass of H_2O=1.963g

Mass of sample = 2.001 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

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

So, in 5.995 g of carbon dioxide, \frac{12}{44}\times 5.995=1.635g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

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

So, in 1.963 g of water, \frac{2}{18}\times 1.963=0.218g of hydrogen will be contained.

Mass of oxygen in the compound = (2.001) - (1.635 + 0.218) = 0.148 g

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

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.635g}{12g/mole}=0.136moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.218g}{1g/mole}=0.218moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.148g}{16g/mole}=0.0092moles

  • <u>Step 2:</u> 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 0.0092 moles.

For Carbon = \frac{0.136}{0.0092}=14.78\approx 15

For Hydrogen = \frac{0.218}{0.0092}=23.69\approx 24

For Oxygen = \frac{0.0092}{0.0092}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 15 : 24 : 1

Hence, the empirical formula for the given compound is C_{15}H_{24}O_1

3 0
4 years ago
More than ½ of the world's forests have been lost to deforestation. True False
RideAnS [48]
The answer is false
8 0
4 years ago
Read 2 more answers
Under the right conditions aluminum will react with chlorine to produce aluminum chloride.
salantis [7]

Answer:

m_{Al}=9.42gAl

Explanation:

Hello there!

In this case, according to the given chemical reaction:

2 Al + 3 Cl2 --> 2 AlCl3

Whereas there is a 2:3 mole ratio of aluminum to chlorine; it will be possible for us to calculate the required grams of aluminum by using the equality 22.4 L = 1 mol, the aforementioned mole ratio and the atomic mass of aluminum (27.0 g/mol) to obtain:

m_{Al}=11.727LCl_2*\frac{1molCl_2}{22.4LCl_2}*\frac{2molAl}{3molCl_2}  *\frac{27.0gAl}{1molAl} \\\\m_{Al}=9.42gAl

Regards!

8 0
3 years ago
Determine the freezing point of 0.368 kg of H2O with 11.85 g of C2H5OH where the kf is 1.86 C/m.
saveliy_v [14]
We will use the formula for freezing point depression :

but first, we need to get the molality m of the solution:

- molality m = moles of C2H5OH / mass of water Kg

 when moles of C2H5OH = mass of C2H5OH/ molar mass of C2H5OH

                                            = 11.85 g / 46 g/mol
                                 
                                           = 0.258 moles

and when we have the mass of water Kg = 0.368 Kg

so, by substitution on the molality formula:

∴ molality m = 0.258 moles / 0.368 Kg

                     = 0.7 mol/Kg

and when C2H5OH is a weak acid so, there is no dissociation ∴ i = 1

and when Kf is given = 1.86 C/m

so by substitution on ΔTf formula:

when ΔTf = i Kf m

∴ ΔTf = 1 * 1.86C/m * 0.7mol/Kg

          = 1.302 °C

5 0
3 years ago
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