CHEMISTRY HIGH SCHOOL

What is the concentration of O2(g), in parts per million, in a solution that contains 0.008 gram of O2(g) dissolved in 1000. grams of H2O(l)?

Answers

Answer 1

Answer:

concentration of O₂ dissolved in H₂O can be written as ppm

ppm stands for parts per million - mg/kg

the amount of mg in 1 kg of solution

the mass of O₂ - 0.008 g

1000 mg equivalent to 1 g

therefore mass of O₂ in mg - 0.008 g x 1000 mg/g = 8 mg

the mass of water is 1000 g

1000 g is equivalent to 1 kg

mass of water in kg - 1000 g / 1000 g/kg = 1 kg

there's 8 mg of O₂ in 1 kg of water

therefore concentration of O₂ is - 8 mg/kg

also can be written as 8 ppm

answer is 8 ppm

Answer 2

Answer:

Answer : The concentration of $O2(g)$ in parts per million is, 8 ppm

Solution : Given,

Mass of oxygen gas (solute) = 0.008 g

Mass of water (solvent) = 1000 g

First we have to calculate the mass of solution.

Mass of solution = Mass of solute + Mass of solvent = 0.008 + 1000 = 1000.008 g

Now we have to calculate the concentration of $O2(g)$ in parts per million.

ppm : It is defined as the mass of solute present in one million $(10^6)$ parts by mass of the solution.

$ppm=\frac{\text{Mass of solute}}{\text{Mass of solution}}* 10^6$

Now put all the given values in this expression, we get

$ppm=(0.008g)/(1000.008g)* 10^6=7.99=8ppm$

Therefore, the concentration of $O2(g)$ in parts per million is, 8 ppm

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What is the affect of increasing the water's mass?how does it reflect it's temperature?

Answers

I assume what you're asking about is, how does the temperature changes when we increase water's mass, according the formula for heat ?
Well the formula is : $Q=m\cdot c\cdot \Delta t$ (where Q is heat, m is mass, c is specific heat and $\Delta t$ is change in temperature. So according this formula, increasing mass will increase the substance's heat, but won't effect it's temperature since they are not related. Unless, if you want to keep the substance's heat constant, in that case when you increase it's mass you will have to decrease the temperature

A chemical reaction in which compounds break up into simpler constituents is a _______ reaction.a. combination
b. decomposition
c. double replacement
d. substitution

Answers

the awnser to ur question is B

B. Decomposition, I remember reading this in my student workbook. Hope this helped!

Choose the statement that best describes the titration from which this data was collected

Answers

I think the correct answer is the first option. The analyte was an acid but and the titrant is a base since from the graph the initial pH of the solution is at an acidic pH when there is still no titrant added.

in a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answers

The partial pressure exerted by a gas in a mixture, depends on the mole

fraction of the gas and the pressure exerted by the mixture.

The partial pressure of H₂O is 20 atm.

Reasons:

Given parameters are;

Explosion equation is 4C₃H₅N₃O₉ → 12CO₂(g) + O₂(g) + 6N₂(g) + 10H₂O(g)

Amount of nitroglycerine = 227 g

Molar mass of nitroglycerine = 227 g/mol

Required:

Partial pressure of the water vapor

Solution:

Number of moles of nitroglycerine in the reaction = 1 mole

Therefore;

Number of moles of CO₂ = 12/4 = 3 moles

Number of moles of O₂ = 0.25 moles

Number of nitrogen, N = 1.5 moles

Number of moles of H₂O = 2.5 moles

$Mole \ fraction \ of \ H_2O, \ X_(H_2O) = (2.5)/(2.5 + 1.5 + 0.25 + 3) = (10)/(29)$

According to Raoults law, we have;

The partial pressure of H₂O = $X_(H_2O) * P_$

Therefore, partial pressure of H₂O = $(10)/(29) * 58$ = 20 atm.

Learn more here:

brainly.com/question/10165688

This is an incomplete question, here is a complete question.

Nitroglycerine (C₃H₅N₃O₉) explodes with tremendous force due to the numerous gaseous products. The equation for the explosion of Nitroglycerine is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

A scientist conducts an experiment to characterize a bomb containing nitroglycerine. She uses a steel, ridge container for the test.

Volume of rigid steel container: 1.00 L

Molar mass of Nitroglycerine: 227 g/mol

Temperature: 300 K

Amount of Nitroglycerine tested: 227 g

Value for ideal gas constant, R: 0.0821 L.atm/mol.K

In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answer : The partial pressure of the water vapor is, 20.01 atm

Explanation :

First we have to calculate the moles of $C_3H_5N_3O_9$

$\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=(227g)/(227g/mol)=1mol$

Now we have to calculate the moles of $CO_2,O_2,N_2\text{ and }H_2O$

The balanced chemical reaction is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

From the balanced chemical reaction we conclude that,

As, 4 moles of $C_3H_5N_3O_9$ react to give 12 moles of $CO_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $(12)/(4)=3$ moles of $CO_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 1 moles of $O_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $(1)/(4)=0.25$ moles of $O_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 6 moles of $N_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $(6)/(4)=1.5$ moles of $N_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 10 moles of $H_2O$

So, 1 moles of $C_3H_5N_3O_9$ react to give $(10)/(4)=2.5$ moles of $H_2O$

Now we have to calculate the mole fraction of water.

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}$

$\text{Mole fraction of }H_2O=(2.5)/(2.5+3+0.25+1.5)=0.345$

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

$p_(H_2O)=X_(H_2O)* p_T$

where,

$p_(H_2O)$ = partial pressure of water vapor gas = ?

$p_T$ = total pressure of gas = 58 atm

$X_(H_2O)$ = mole fraction of water vapor gas = 0.345

Now put all the given values in the above formula, we get:

$p_(H_2O)=X_(H_2O)* p_T$

$p_(H_2O)=0.345* 58atm=20.01atm$

Therefore, the partial pressure of the water vapor is, 20.01 atm

Carbon forms molecular compounds with some elements from Group 16. Two of these compounds are carbon dioxide, CO2, and carbon disulfide, CS2.Carbon dioxide is a colorless, odorless gas at room temperature. At standard temperature and pressure, CO2(s) changes directly to CO2(g).

Carbon disulfide is formed by a direct reaction of carbon and sulfur. At room temperature, CS2 is a colorless liquid with an offensive odor. Carbon disulfide vapors are flammable.

59 Identify one physical property and one chemical property of CS2. [1]

Answers

Answer: Physical property of $CS_2$ is state of matter (liquid) and it is a chemical compound.

Explanation:

Chemical property is defined as the property of a substance which is observed during a reaction where the chemical composition identity of the substance gets changed.

Physical property is defined as the property which can be measured and whose value describes the state of physical system. For Example: State, density etc.

Carbon disulfide $(CS_2)$ is the compound which is formed by the chemical combination of carbon and sulfur atoms. The equation for this follows:

$C+2S\rightarrow CS_2$

The chemical property of carbon disulfide is different from that of carbon and sulfur. Thus, this is a chemical compound.

It is given, that this compound is a colorless liquid, which means that the state of this compound is liquid state.

Hence, physical property of $CS_2$ is state of matter (liquid) and it is a chemical compound.

The physical property of carbon disulfide is that at room temperature, CS2 is a colorless liquid with an offensive odor. The chemical property of carbon disulfide is that it is formed by a direct reaction of carbon and sulfur.

Which is the correctly balanced chemical equation for the reaction of KOH and H2SO4?

Answers

The answer should be 2KOH+H2SO4 ---> K2SO4+2H2O.
I hope this helps!!

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