Help asap please 1!!!!

Is layers of sediment forming at the bottom of the ocean erosion?

kiruha [24]

Yes layers of sediment form at the bottom of the ocean.

5 0

3 years ago

Is burning of coal a reversible change??? If yes how?

koban [17]

I don't think so. No way that I know anyway. It it could be done then the need for more coal to be mined would have stopped hundreds of years ago. Once coal is burned, it forms water and carbon dioxide (essentially) with some sulfur oxides.

How do you put that back together again. It's a little like humpty dumpty.

5 0

3 years ago

Read 2 more answers

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

dimulka [17.4K]

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 $C_3H_5N_3O_9$

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

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 $\frac{12}{4}=3$ 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 $\frac{1}{4}=0.25$ 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 $\frac{6}{4}=1.5$ 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 $\frac{10}{4}=2.5$ moles of $H_2O$

Now we have to calculate the mole fraction of water.

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

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

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

According to the Raoult's law,

$p_{H_2O}=X_{H_2O}\times p_T$

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:

$p_{H_2O}=X_{H_2O}\times p_T$

$p_{H_2O}=0.345\times 58atm=20.01atm$

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

3 0

3 years ago

A solution is made by dissolving 10.7 g of magnesium sulfate, MgSO4, in enough water to make exactly 100 mL of solution. Calcula

adell [148]

Answer:

[MgSO₄] = 890 mM/L

Explanation:

In order to determine molarity we need to determine the moles of solute that are in 1L of solution.

Solute: MgSO₄ (10.7 g)

Solvent: water

Solution: 100 mL as volume. (100 mL . 1L / 1000mL) = 0.1L

We convert the solute's mass to moles → 10.7 g / 120.36 g/mol = 0.089 moles

Molarity (mol/L) → 0.089 mol/0.1L = 0.89 M

In order to calculate M to mM/L, we make this conversion:

0.89 mol . 1000 mmoles/ 1 mol = 890 mmoles

4 0

3 years ago

Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

Anna11 [10]

Answer:

[COF₂] = 0.346M

Explanation:

For the reaction:

2COF₂(g) ⇌ CO₂(g) + CF₄(g)

Kc = 5.70 is defined as:

Kc = [CO₂] [CF₄] / [COF₂]² = 5.70 (1)

Equilibrium concentrations of each compound after addition of 2.00M COF₂ will be:

[COF₂] : 2.00M - 2x

[CO₂] : x

[CF₄] : x

Replacing in (1):

5.70 = [X] [X] / [2-2x]²

22.8 - 45.6x + 22.8x² = x²

0 = -21.8x² + 45.6x - 22.8

Solving for x:

X = 1.265 -False answer, will produce negative concentrations-

X = 0.827.

Replaing, molar concentration of COF₂ is:

[COF₂] : 2.00M - 2×0.827 = 0.346M

I hope it helps!

7 0

3 years ago