All STEM subjects require ________________ skills. a. Same b. Different

A sample of gas at 47C and 1.5 pressure occupies a volume of 2.20L. What volume would this gas occupy at 107C and 2.5 pressure?

Brums [2.3K]

Answer:

Explanation:

21

8 0

3 years ago

How many moles of water can be produced with 4.3 moles of H2 and 5.6 moles of O2? Which reactant is limiting? How many moles of

xeze [42]

Answer:

Hydrogen H₂ will be the limiting reagent.

The excess reactant that will be left after the reaction is 3.45 moles.

4.3 moles of water can be produced.

Explanation:

The balanced reation is:

2 H₂ + O₂ → 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

H₂: 2 moles
O₂: 1 mole
H₂O: 2 moles

To determine the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 1 mole of O₂ reacts with 2 moles of H₂, how much moles of H₂ will be needed if 5.6 moles of O₂ react?

$moles of H_{2} =\frac{5.6 moles of O_{2} *2 mole of H_{2} }{1 mole of O_{2}}$

moles of H₂= 11.2 moles

But 11.2 moles of H₂ are not available, 4.3 moles are available. Since you have less moles than you need to react with 5.6 moles of O₂, hydrogen H₂ will be the limiting reagent and oxygen O₂ will be the excess reagent.

Then you can apply the following rules of three:

If by reaction stoichiometry 2 moles of H₂ react with 1 mole of O₂, 4.3 moles of H₂ will react with how many moles of O₂?

$moles of O_{2} =\frac{1 mole of O_{2} *4.3 mole of H_{2} }{2 mole of O_{2}}$

moles of O₂= 2.15 moles

The excess reactant that will be left after the reaction can be calculated as:

5.6 moles - 2.15 moles= 3.45 moles

The excess reactant that will be left after the reaction is 3.45 moles.

If by reaction stoichiometry 2 moles of H₂ produce 2 moles of H₂O, 4.3 moles of H₂ produce how many moles of H₂O?

$moles of H_{2}O =\frac{2 moles of H_{2}O *4.3 mole of H_{2} }{2 mole of H_{2}}$

moles of H₂O= 4.3 moles

4.3 moles of water can be produced.

8 0

3 years ago

I’ll give brainliest please help thanks.

oksano4ka [1.4K]

Answer:

See Explanation

Explanation:

The equation of the reaction is;

C7H16(g) + 11O2(g) ---->7CO2(g) + 8H2O(g)

1a) Number of moles in 228 g of H2O = 228 g/18 g/mol = 12.67 moles

From the reaction equation;

11 moles of O2 yields 8 moles of H2O

x moles of O2 yields 12.67 moles of H2O

x = 11 * 12.67/8

x = 17.4 moles of O2

Since 1 mole of O2 occupies 22.4 L

17.4 moles of O2 occupies 17.4 moles * 22.4 L/1 mole = 389.76 L

1b) Molar mass of C7H16 = 100 g/mol

Number of moles in 300 g of C7H16 = 300 g/100g/mol = 3 moles

1 mole of C7H16 yields 7 moles of CO2

3 moles of C7H16 yields 3 * 7/1 = 21 moles of CO2

If 1 mole of CO2 occupies 22.4 L

21 moles of CO2 occupies 21 moles * 22.4 L/1 mole = 470.4 L of CO2

2) Number of moles in 202 L of H2 is obtained by;

1 mole of H2 occupies 22.4 L

x moles of H2 occupies 202 L

x = 1 mole * 202 L/22.4 L = 9 moles of H2

From the reaction equation;

3 moles of H2 yields 2 moles of PH3

9 moles of H2 will yield 9 * 2/3 = 6 moles of PH3

Mass of 6 moles of PH3 = 6 moles of PH3 * 34 g/mol = 204 g of PH3

3 0

3 years ago

Air, with an initial absolute humidity of 0.016 kg water per kg dry air, is to be dehumidified for a drying process by an air co

amid [387]

Answer:

(a) The proportion of dry air bypassing the unit is 14.3%.

(b) The mass of water removed is 1.2 kg per 100 kg of dry air.

Explanation:

We can express the proportion of air that goes trough the air conditioning unit as $p_{d}$ and the proportion of air that is by-passed as $p_{bp}$ , being $p_{d}+p_{bp}=1$ .

The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:

$0.004 = 0.016*p_{bp}+ 0.002*p_{d}$

Replacing the first equation in the second one we have

$0.004 = 0.016*(1-p_{d})+ 0.002*p_{d}=0.016-0.016*p_{d}+0.002*p_{d}\\0.004 - 0.016 = (-0.016+0.002)*p_{d}\\-0.012 = -0.014*p_{d}\\p_{d}=\frac{-0.012}{-0.014}=0.857\\\\p_{bp}=1-p_{d}=1-0.857=0.143$

(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.

It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.

The water removed of every 100 kg of dry air is

$100 kgDA*0.857*0.014 kgW/kgDA= 1.1998 \approx 1.2 kgW$

It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).

100 * 0.012 = 1.2 kgW

7 0

3 years ago

Which is the formula for dinitrogen pentoxide? A)N5O2 B)5N2O C) 2NO5 D) N2O5

kompoz [17]

Answer is d.........

8 0

3 years ago

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