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3 years ago

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The compound consists of 40.1% carbon, 6.6% hydrogen and 53.3% oxygen. Its relative density with respect to hydrogen is -15. Wha

t is the formula of the compound?

1 answer:

kap26 [50]3 years ago

3 0

Answer:

https://socratic.org/answers/220339

this is the answer , my grandpa create this page

Explanation:

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The graph shows the speed of a car

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Answer:

Letter B.

Explanation:

Changing its direction and climbing a steep incline can't be proven only by the graph info. You can see that it's speed isn't constant in the graph.

We have letter B left.

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3 years ago

after a chemistry student made a AgNO3(small 3 at bottom) solution., she wanted to determine the molar concentration of it. if 2

Alexxandr [17]

Molar concentration = (numbet of mol Solute)/ ( volume Solution)

1) Finding
the number of the mol solute

$22.0 g AgNO3 * \frac{1 mol AgNO3}{169.91 g AgNO3} = 0.129 mol AgNO3

2) 725 ml = 0.725 L


$

$3) Molarity = \frac{number/ mol/solute}{Volume/solution} = \frac{0.129}{0.725} 

$

$Molarity= \frac{0.178 mol}{L} , 

or 0.178 M
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3 years ago

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3 years ago

Read 2 more answers

The acid HOCl (hypochlorous acid) is produced by bubbling chlorine gas through a suspension of solid mercury(II) oxide particles

sergejj [24]

<u>Answer:</u> The expression for equilibrium constant is $K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}$

<u>Explanation:</u>

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{eq}$

For the general chemical equation:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_c$ is given as:

$K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

For the given chemical reaction:

$2HgO(s)+H_2O(l)+2Cl_2(g)\rightleftharpoons 2HOCl(aq.)+HgO.HgCl_2(s)$

The expression for $K_{eq}$ is given as:

$K_{eq}=\frac{[HOCl]^2[HgO.HgCl_2]}{[HgO]^2[H_2O][Cl_2]^2}$

The concentration of solid is taken to be 0.

So, the expression for $K_{eq}$ is given as:

$K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}$

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

5 0

3 years ago