Ask question

Ask question

All categories

3 years ago

15

The Henry's law constant for helium gas in water at 30 °C is 3.70 × 10-4 M/atm. When the partial pressure of helium above a samp

le of water is 0.650 atm, the concentration of helium in the water is __________ M

1 answer:

igor_vitrenko [27]3 years ago

4 0

Answer:

The concentration of helium in the water is 2.405×10^-4 M

Explanation:

Concentration = Henry's law constant × partial pressure of helium

Henry's law constant = 3.7×10^-4 M/atm

Partial pressure of helium = 0.65 atm

Concentration = 3.7×10^-4 × 0.65 = 2.405×10^-4 M

You might be interested in

H2SO4 + 2NaOH → 2H2O + Na So.

gayaneshka [121]

Answer:

2 mol H

Explanation:

For every 2 mol of NaOH, we're reacting 2 mol of H2O. In order to figure out how many mol of H are needed, it needs to be set up stochiometrically. Starting off with the given value, 1 mol of NaOH, we can then make a mol to mol ratio. For 2 mol of NaOH, we have 2 mol of H2O. For every 2 mol of H2O, we have 4 mol of H (this is because we are multiplying the coefficient by the subscript: 2 × 2). Now, we can solve for our answer.

1 mol NaOH × (2 mol H₂O / 2 mol NaOH) × (4 mol H / 2 mol H₂O)

= 2 mol H

Thus, we get 2 mol of H are needed to completely react 1 mol of NaOH.

7 0

4 years ago

A) Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of CO2 and 1.37 mg of H2O. If the compound contains o

Mumz [18]

<u>Answer:</u>

<u>For a:</u> The empirical formula for the given compound is $CH$

<u>For b:</u> The empirical and molecular formula for the given organic compound are $C_{10}H_{20}O$

<u>Explanation:</u>

<u>For a:</u>

The chemical equation for the combustion of hydrocarbon follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x', and 'y' are the subscripts of Carbon and hydrogen respectively.

We are given:

Conversion factor used: 1 g = 1000 mg

Mass of $CO_2=5.86mg=5.86\times 10^{-3}g$

Mass of $H_2O=1.37mg=1.37\times 10^{-3}g$

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 $5.86\times 10^{-3}g$ of carbon dioxide, $\frac{12}{44}\times 5.86\times 10^{-3}=1.60\times 10^{-3}g$ 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 $1.37\times 10^{-3}g$ of water, $\frac{2}{18}\times 1.37\times 10^{-3}=0.152\times 10^{-3}g$ of hydrogen will be contained.

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

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

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.60\times 10^{-3}g}{12g/mole}=0.133\times 10^{-3}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.152\times 10^{-3}g}{1g/mole}=0.152\times 10^{-3}moles$

<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 $0.133\times 10^{-3}$ moles.

For Carbon = $\frac{0.133\times 10^{-3}}{0.133\times 10^{-3}}=1$

For Hydrogen = $\frac{0.152\times 10^{-3}}{0.133\times 10^{-3}}=1.14\approx 1$

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

The ratio of C : H = 1 : 1

Hence, the empirical formula for the given compound is $CH$

<u>For b:</u>

The chemical equation for the combustion of menthol 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 $CO_2$ = 0.2829 g

Mass of $H_2O$ = 0.1159 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 0.2829 g of carbon dioxide, $\frac{12}{44}\times 0.2829=0.077g$ 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 0.1159 g of water, $\frac{2}{18}\times 0.1159=0.013g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1005) - (0.077 + 0.013) = 0.105 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 = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.013g}{1g/mole}=0.013moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0105g}{16g/mole}=0.00065moles$

<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 0.00065 moles.

For Carbon = $\frac{0.0064}{0.00065}=9.84\approx 10$

For Hydrogen = $\frac{0.013}{0.00065}=20$

For Oxygen = $\frac{0.00065}{0.00065}=1$

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

The ratio of C : H : O = 10 : 20 : 1

The empirical formula for the given compound is $C_{10}H_{20}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 156 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

$n=\frac{156g/mol}{156g/mol}=1$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O$

Hence, the empirical and molecular formula for the given organic compound are $C_{10}H_{20}O$

3 0

3 years ago

A substance that contains hydrogen that may be replaced by a metal is a/an

vovangra [49]

The substance that contains hydrogen that may be replaced by a metal is an acid~

3 0

3 years ago

Read 2 more answers

If 309 grams of potassium combines with 228 g chlorine gas, how many grams of the single product will be formed? What are the LR

lisov135 [29]

The amount of KCl that would be formed will be 478.8 grams

<h3>Stoichiometric calculations</h3>

From the equation of the reaction, the mole ratio of K to $Cl_2$ is 2:1.

Mole of 309 g potassium = 309/39 = 7.92 moles

Mole of 228 g Cl2 = 228/71 = 3.21 moles

Thus, potassium is in excess while Cl2 is limiting.

Mole ratio of Cl2 and KCl = 1:2

Equivalent mole of KCl = 3.21 x 2 = 6.42 moles

Mass of 6.42 moles KCl = 6.42 x 74.55 = 478.8 grams

More on stoichiometric calculations can be found here: brainly.com/question/27287858

#SPJ1

4 0

2 years ago

1)Substance A melts at 2300 °c, and it doesn’t conduct electricity even when molten and doesn't dissolve in water. What is the s

Minchanka [31]

Answer:
Substance A is a Giant covalent structure with covalent bonding

Explanation:
Substance A is a Giant covalent structure. There are strong covalent bonds between the atoms and a large amount of heat energy is needed to break the bonds resulting in high melting point. Substance A does not conduct electricity in molten state reason being, it exist as molecules which are electrically neutral and there are no mobile charged carriers.

Hope this helps!

5 0

2 years ago