Which statement describes the best way to determine how different levels of light affect the growth of seedling plants?

If 588 grams of FeS2 is allowed to react with 352 grams of O2 according to the following equation, how many grams of Fe2O3 are p

Rainbow [258]

Answer: The mass of iron (III) oxide produced is 782.5 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $FeS_2$ :

Given mass of $FeS_2$ = 588 g

Molar mass of $FeS_2$ = 120 g/mol

Putting values in equation 1, we get:

$\text{Moles of }FeS_2=\frac{588g}{120g/mol}=4.9mol$

For $O_2$ :

Given mass of $O_2$ = 352 g

Molar mass of $O_2$ = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of }O_2=\frac{352g}{32g/mol}=11mol$

The chemical equation for the reaction of $FeS_2$ and oxygen gas follows:

$FeS_2+O_2\rightarrow Fe_2O_3+SO_2$

By Stoichiometry of the reaction:

1 mole of $FeS_2$ reacts with 1 mole of oxygen gas

So, 4.9 moles of $FeS_2$ will react with = $\frac{1}{1}\times 4.9=4.9mol$ of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, $FeS_2$ is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of $FeS_2$ produces 1 mole of iron (III) oxide

So, 4.9 moles of $FeS_2$ will produce = $\frac{1}{1}\times 4.9=4.9moles$ of iron (III) oxide

Now, calculating the mass of iron (III) oxide from equation 1, we get:

Molar mass of iron (III) oxide = 159.7 g/mol

Moles of iron (III) oxide = 4.9 moles

Putting values in equation 1, we get:

$4.9mol=\frac{\text{Mass of iron (III) oxide}}{159.7g/mol}\\\\\text{Mass of iron (III) oxide}=(4.9mol\times 159.7g/mol)=782.5g$

Hence, the mass of iron (III) oxide produced is 782.5 grams

8 0

3 years ago

What can help to stop soil erosion ?

DanielleElmas [232]

Answer:

just read the explanation.

Explanation:

1.Maintaining a healthy, perennial plant cover.

2.Mulching.

3.Planting a cover crop – such as winter rye in vegetable gardens.

4.Placing crushed stone, wood chips, and other similar materials in heavily used areas where vegetation is hard to establish and maintain.

Hope this helps. :)

4 0

3 years ago

The discharge of chromate ions (CrO42-) to sewers or natural waters is of concern because of both its ecological impacts and its

muminat

Answer:

45727g

Explanation:

So, have the overall ionic equation given as the following;

CrO42^- + 3 Fe2^+ + 8 H2O ------> Cr(OH)^3(s) + 3 Fe(OH)^3(s) + 4 H^+.

So, we have (from the question) that the amount or quantity of the waste stream daily = 60m^3/h, and the waste stream daily contains waste stream containing = 4.0 mg/L Cr, and the discharge limit = 0.1 mg/L.

Step one: convert m^3/ h to L/h. Therefore, 60 m^3/h × 1000dm^3 = 60000 L/h .

Step two: Determine or calculate the the value of Cr used up.

The value of Car used up ={ 60,000 × ( 4.0 - 0.1) } ÷ 1000 = 234 g.

Step three: Determine or calculate the mass of Cr(OH)3 and the mass of Fe(OH)3.

The number of moles of Cr = 234/52 = 4.5 moles.

Molar mass of Cr(OH)3 = 103 g/mol and the molar mass of Fe(OH)3 = 106.8 g/mol.

Thus, the mass of Cr(OH)3 = 4.5 × 103 = 463.5 g.

And the mass of Fe(OH)3 = 13.5 × 106.8 = 1441.8 g.

Hence, the total = 463.5 g + 1441.8 g = 1905.3 g.

Step four: Determine or calculate the How much particulate matter would be generated daily.

The amount of the particulate that would be generated daily = 24 × 1905.3 = 45727g.

7 0

3 years ago

For the reaction of oxygen and nitrogen to form nitric oxide, consider the following thermodynamic data: ΔH∘rxn 180.5 kJ/mol ΔS∘

nasty-shy [4]

Answer:

7.28 × 10³ K

Explanation:

Let's consider the following reaction.

N₂(g) + O₂(g) → 2 NO(g)

The reaction is spontaneous when the standard Gibbs free energy (ΔG°) is negative. ΔG° is related to the standard enthalpy of the reaction (ΔH°) and the standard entropy of the reaction (ΔS°) through the following expression.

ΔG° = ΔH° - T . ΔS°

If ΔG° < 0,

ΔH° - T . ΔS° < 0

ΔH° < T . ΔS°

T > ΔH°/ΔS° = (180.5 × 10³ J/mol)/(24.8 J/mol.K) = 7.28 × 10³ K

The reaction is spontaneous above 7.28 × 10³ K.

7 0

4 years ago

The heat capacity of air is much smaller than that of water, and relatively modest amounts of heat are needed to change its temp

LekaFEV [45]

Answer:

$Q=9.2x10^5J$

$t=614s=10.2min$

Explanation:

Hello,

In this case, we can compute the energy by using the following formula for air:

$Q=nCp\Delta T$

Whereas the moles of air are computed via the ideal gas equation at room temperature inside the 5.5m x 6.5m x 3.0m-room:

$n=\frac{PV}{RT}\\\\V=5.5m*6.5m*3.0m=107.25m^3*\frac{1000L}{1m^3}=107250L\\ \\n=\frac{1atm*107250L}{0.082\frac{atm*L}{mol*K}*298.15K}\\ \\n=4386.8mol$

Now, we are able to compute heat, by considering that the temperature raise is given in degree Celsius or Kelvins as well:

$Q=4386.8mol*21\frac{K}{mol*K}*10K \\\\Q=9.2x10^5J$

Finally, we compute the time required for the heating by considering the heating rate and the required heat, shown below:

$t=\frac{9.2x10^5J}{1.5\frac{kJ}{s}*\frac{1000J}{1kJ} } \\\\t=614s=10.2min$

Regards.

8 0

3 years ago