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3 years ago

Molar mass of SeOBr2 with work (PLEASE HELP ME!)

Chemistry

2 answers:

iogann1982 [59]3 years ago

8 0

Answer:

254.77 g/mol

I don't know what you mean by "with work"

Explanation:

The molar mass is the mass of a given chemical element or chemical compound (g) divided by the amount of substance (mol).

Trava [24]3 years ago

3 0

Answer:

$\boxed {\boxed {\sf 254.679 \ g/mol }}$

Explanation:

The molar mass is the mass of 1 mole of particles in a substance. We can calculate using values on the Periodic Table.

First, identify the elements in the compound: SeOBr₂.

Selenium (Se), Oxygen (O), and Bromine (Br)

Use the Periodic Table to find the molar masses.

Se: 78.97 g/mol
O: 15.999 g/mol
Br: 79.90 g/mol

Notice the formula has a subscript of 2 after bromine. This means there are 2 moles of bromine in 1 mole of SeOBr₂.

To calculate the molar masses, multiply bromine's molar mass by 2, then add selenium and oxygen.

$2(79.90 \ g/mol) + 15.999 \ g/mol + 78.97 \ g/mol \\159.80 \ g/mol + 15.999 \ g/mol + 78.97 \ g/mol \\254.679 \ g/mol$

The molar mass of SeOBr₂ is <u>254.679 grams per mole</u>.

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How many hydrogen atoms are in 89.5 g of<br> C6H6 ?<br> Answer in units of atoms.

Elodia [21]

Solution :

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So, 89.5 gram of $C_6H_6$ contains :

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Now, from the formula we can see that one molecule of $C_6H_6$ contains 2 hydrogen atom . So, number of hydrogen atom are :

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3 years ago

Given the following data:

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
$\text{NH}_3\; (g)$ : $-2 \; b = -1$
$\text{HCl} \; (g)$ : $2 \; c = -1$

Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

$a + b = -1/2 + 1/2 = 0$

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

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3 0

3 years ago

The combustion of 40.10 g of a compound which contains only C, H, Cl and O yields 58.57 g of CO2 and 14.98 g of H2O. Another sam

elena-14-01-66 [18.8K]

Answer:

The empirical formula is C4H5ClO2

Explanation:

Step 1: Data given

Mass of the sample = 40.10 grams

Mass of CO2 produced = 58.57 grams

Mass of H2O produced = 14.98 grams

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Atomic mass C= 12.01 g/mol

Atomic mass O = 16.0 g/mol

Atomic mass H = 1.01 g/mol

In experiment 2, mass = 75.00 grams and 22.06 grams is Cl

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 58.57 grams / 44.01 g/mol

Moles CO2 = 1.33 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol CO2

For 1.33 moles CO2 we have 1.33 moles C

Step 4: Calculate mass C

Mass C = 1.33 grams * 12.01 g/mol

Mass C = 15.97 grams

Step 5: Calculate moles H2O

Moles H2O= 14.98 grams /18.02 g/mol

Moles H2O = 0.831 moles

Step 6: Calculate moles H

For 1mol H2O we have 2 moles H

For 0.831 moles H2O we have 2*0.831 = 1.662 moles H

Step7: Calculate mass H

Mass H = 1.662 moles * 1.01 g/mol

Mass H = 1.68 grams

Step 8: Calculate mass %

%C = (15.97 grams / 40.10) * 100 %

%C = 39.8 %

%H = (1.68 / 40.10 ) *100%

%H = 4.2 %

%Cl = (22.06 / 75.00 ) * 100%

%Cl = 29.4 %

%O = 100 % - 39.8% - 4.2 % - 29.4 %

%O = 26.6 %

Step 9: Calculate moles in compound

We assume the compound has a mass of 100 grams

Mass C = 39.8 grams

MAss H = 4.2 grams

MAss Cl = 29.4 grams

Mass O = 26.6 grams

Moles C = 39.8 grams / 12.01 g/mol

Moles C = 3.314 moles

Moles H = 4.2 moles / 1.01 g/mol

Moles H = 4.158 moles

Moles Cl =29.4 grams / 35.45 g/mol

Moles Cl = 0.829 moles

Moles O = 26.6 grams / 16.0 g/mol

Moles O = 1.663 moles

Step 10: calculate the mol ratio

We divide by the smallest amount of moles

C: 3.314 moles / 0.829 moles = 4

H: 4.158 moles / 0.829 moles = 5

Cl: 0.829 moles /0.829 moles = 1

O: 1.663 moles / 0.829 moles = 2

This means For each Cl atom we have 4 C atoms, 5 H atoms and 2 O atoms

The empirical formula is C4H5ClO2

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3 years ago

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konstantin123 [22]

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