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Stells [14]
3 years ago
14

The concentration of CI ion in a sample of H,0 is 15.0 ppm. What mass of CI ion is present in 240.0 mL of H,0, which has a densi

ty of 1.00 g/mL?
Chemistry
1 answer:
Doss [256]3 years ago
8 0

Answer:

Mass of solute = 0.0036 g

Explanation:

Given data:

Concentration of Cl⁻ = 15.0 ppm

Volume of water = 240 mL

Mass of Cl⁻ present = ?

Solution:

1 mL = 1 g

240 mL = 240 g

Formula:

ppm = mass of solute / mass of sample ×1,000,000

by putting values,

15.0 ppm = (mass of solute / 240 g) ×1,000,000

Mass of solute = 15.0 ppm ×  240 g / 1,000,000

Mass of solute = 0.0036 g

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The molar mass of hydrogen molecules is 2g/mol and the molar mass of oxygen molecules is 32g/mol. What is the ratio of the avera
Alona [7]

Answer:

1

Explanation:

For an ideal gas, the average kinetic energy is given by:

Ek = (3/2)*n*R*T

Where n is the number of moles, R is the gas constant (8.31 J/mol*K), and T the temperature. The gases have the same number of moles, and the same temperature, so they will have the same average kinetic energy:

Ek = (3/2)*1*8.31*300

Ek =3739.5 J

So, the ratio between then is 1.

8 0
4 years ago
Would it be C..? <br> More NO2 and SO2 would form..help..please and thank you
Scorpion4ik [409]

Answer:

C. More NO2 and SO2 will form

Explanation:

Le Chatelier's Principle : It  predicts the behavior of equilibrium due to change in pressure , temperature , volume , concentration etc

It states that When external changes are introduced in the equilibrium then it will shift the equilibrium in a direction to reduce the change.

In given Reaction SO3 is introduced(increased) .

So equilibrium will shift in the direction where SO3 should be consumed(decreased)

Hence the equilibrium will go in backward direction , i.e

NO + SO_{3} \rightarrow NO_{2} + SO_{2}

So more and more Of NO2 and SO2 will form

7 0
4 years ago
Which of the following is not capable of reacting with molecular oxygen?
fredd [130]
I think letter b were you put the letters so
6 0
3 years ago
Here is a more complex redox reaction involving the dichromate ion in acidic solution: 3N O 2 − + 8H + + C r 2 O 7 2− → 3N O 3 −
Fantom [35]

Answer:

NO2- is the reducing agent.

Cr2O7_2- is the oxidizing agent.

H+ is neither

Explanation:

Reduction is the gain in electron. A chemical specie that undergoes reduction is called the oxidizing agent.

Oxidation is simply the loss in electrons. A chemical specie that undergoes oxidation is called the reducing agent.

Let us look at the species.

The first specie is the NO2-. In this specie, the oxidation number of nitrogen changed from +3 to +5 in NO3-. Thus we can see that there is more loss of electron to have caused an increase in the oxidation number positively. This shows an oxidation. Hence, NO2- is the reducing agent.

Let us look at the chromium. We can see that the oxidation number of chromium changed from +7 to +3.

Now we can see that it is a decrease and hence, it is a gain of electron and thus it is reduction. This means the first chromium specie is the oxidizing agent.

The hydrogen ion is simply placed there to balance the ions and hence it is neither the oxidizing nor the reducing agent.

4 0
4 years ago
32 g of sulfur will react with 48 g of oxygen to produce 80 g of sulfur trioxide. If 32 g of sulfur and 100 g of oxygen are plac
Lina20 [59]

Answer:

Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)

Explanation:

Step 1: Data given

Mass of sulfur = 32.00 grams

Mass of oxygen = 48.00 grams

Molar mass of sulfur = 32.07 g/mol

Molar mass of oxygen = 32 g/mol

Molar mass of SO3 = 80.07 g/mol

Step 2: The balanced equation

2S + 3O2 → 2SO3

Step 3: Calculate moles S

Moles S = Mass S / molar mass S

Moles S = 32.0 grams / 32.07 g/mol

Moles S = 0.998 moles

Step 4: Calculate moles O2

Moles O2 = 100.0 grams / 32.0 g/mol

Moles O2 = 3.125 moles

Step 5: Calculate the limiting reactant

For 2 moles S we need 3 moles O2 to produce 2 moles SO3

S is the limiting reactant. It will completely be consumed (0.998 moles)

O2 is in excess, there will be consumed 3/2 * 0.998 = 1.497 moles

There will remain 3.125- 1.497 = 1.628 moles O2

This is 1.628 moles * 32 g/mol = 52.1 grams

Step 6: Calculate moles SO3

For 2 moles S we need 3 moles O2 to produce 2 moles SO3

For 0.998 moles S there will react 0.998 moles SO3

Step 6: Calculate mass SO3

Mass SO3 = moles SO3 * molar mass SO3

Mass SO3 = 0.998 moles * 80.07 g/mol

Mass SO3 = 79.9 grams ≈ 80 grams

There will be produced 80 grams of SO3

Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)

4 0
3 years ago
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