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yanalaym [24]
3 years ago
12

How many grams of iron oxide, Fe2O3 will be produced if 165 g of O2 gas is supplied? (follow the same steps as mol to mol, only

now your flow should be like this: grams O2 moles O2  moles Fe2O3  grams Fe2O3 Fe + O2  Fe2O3
Chemistry
1 answer:
forsale [732]3 years ago
8 0

Answer:

m_{Fe_2O_3}=549gFe_2O_3

Explanation:

Hello there!

In this case, according to the given chemical reaction for this problem about stoichiometry:

4Fe+3O_2\rightarrow 2Fe_2O_3

Whereas there is a 3:2 mole ratio of oxygen (molar mass = 32.0 g/mol) to iron (III) oxide (molar mass = 159.69 g/mol) and therefore, the correct stoichiometric setup is:

m_{Fe_2O_3}=165gO_2*\frac{1molO_2}{32.00gO_2}*\frac{2molFe_2O_3}{3molO_2} *\frac{159.69gFe_2O_3}{1molFe_2O_3}  \\\\m_{Fe_2O_3}=549gFe_2O_3

Regards!

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The volume of carbon gas at 5.0 ATM was measured to be 363 ml. What will the pressure be if the volume is change to .00020 ml?
aleksley [76]

Answer:

The answer to your question is P2 = 9075000 atm

Explanation:

Data

Pressure 1 = P1 = 5 atm

Volume 1 = V1 = 363 ml

Pressure 2 = P2 = ?

Volume 2 = 0.0002 ml

Process

To solve this problem use Boyle's law

                 P1V1 = P2V2

-Solve for P2

                 P2 = P1V1/V2

-Substitution

                  P2 = (5 x 363) / 0.0002

-Simplification

                  P2 = 1815 / 0.0002

-Result

                 P2 = 9075000 atm

5 0
3 years ago
The following diagrams represent mixtures of NO(g) and O2(g). These two substances react as follows: 2NO(g)+O2(g)→2NO2(g) It has
Alja [10]

This is an incomplete question, here is a complete question and an image is attached below.

The following diagrams represent mixtures of NO(g) and O₂(g). These two substances react as follows:

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

It has been determined experimentally that the rate is second order in NO and first order in O₂.

Based on this fact, which of the following mixtures will have the fastest initial rate?

The mixture (1). The mixture (2). The mixture (3).

Answer : The mixture 1 has the fastest initial rate.

Explanation :

The given chemical reaction is:

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

The rate law expression is:

Rate=k[NO]^2[O_2]

Now we have to determine the number of molecules of NO\text{ and }O_2

In mixture 1 : There are 5 NO and 4 O_2 molecules.

In mixture 2 : There are 7 NO and 2 O_2 molecules.

In mixture 3 : There are 3 NO and 5 O_2 molecules.

Now we have to determine the rate law expression for mixture 1, 2 and 3.

The rate law expression for mixture 1 is:

Rate=k[NO]^2[O_2]

Rate=k(5)^2\times (4)

Rate=k(100)

The rate law expression for mixture 2 is:

Rate=k[NO]^2[O_2]

Rate=k(7)^2\times (2)

Rate=k(98)

The rate law expression for mixture 3 is:

Rate=k[NO]^2[O_2]

Rate=k(3)^2\times (5)

Rate=k(45)

Hence, the mixture 1 has the fastest initial rate.

4 0
3 years ago
What is the molarity of a solution containing 0.325 moles of solute in 250 ml of solution?
satela [25.4K]
Not 100%, but I think it would be 0.0013 because the equation for Molarity is Moles of solute(Mol)/ Liters of solution(L)
7 0
3 years ago
When reading any volume in the laboratory, we always read where the bottom of the water curve falls. This is called the______?
zavuch27 [327]

Answer: Meniscus

Explanation: There is a condition on reading volumes. For concave meniscus read volume in the lower part and for convex meniscus read volume at the upper part.

7 0
3 years ago
Read 2 more answers
The gas law for an ideal gas at absolute temperature T (in kelvins), pressure P (in atmospheres), and volume V (in liters) is PV
allsm [11]

Answer:

The rate of change of the temperature is 0.0365 Kelvin per minute.

Explanation:

<u>Step 1</u>: Given data

ideal gas law: P*V = n*R*T

with P= pressure of the gas ( in atm) = 9.0 atm

with V= volume of the gass (in L) =12L

with n = number of moles = 10 moles

R = gas constant = 0.0821 L*atm* K^−1*mo^−1

T = temperature = TO BE DETERMINED

The volume  decreases with a rate of 0.17L/min = dV/dT  = -0.17

The pressure increases at a rate of 0.13atm/min = dP/dT

<u>Step 2:</u> The ideal gas law

P * [dV/dT] + V * [dP/dT] = nR * dT/dt

9 atm * (-0.17L/min) + 12L * 0.13atm/min = 10 moles * 0.0821 L*atm* K^−1*mo^−1 *dT/dt

0.03 = 0.821 * dT/dt

dT/dt = 0.03/0.821

dT/dt = 0.0365

Since the gas constant is expressed in Kelvin and not in °C, this means that <u>the rate of chagnge of the temperature is 0.0365 Kelvin per 1 minute.</u>

7 0
3 years ago
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