What is the mass in grams of H₂ that can be formed from 54.6 grams of NH₃ in the following reaction?2 NH₃(g) → 3 H₂(g) + N₂(g)

Answer:

1. 176 × 10^12 W ; 78400000000

Explanation:

Given the following :

Fall rate = 2,400,000kg/s

Average height of fall = 50m

Gravitational Potential of falling water = mgh = mass × acceleration due to gravity × height =

How many 15 W LED light bulbs could it power?

Recall : power = workdone / time

Workdone = gravitational potential energy

Mass of water = density * volume

Density of water = 1 * 10^3kg/m^3

Rate of fow = volume / time = 2400000

Hence,

Power = 1000 * 2,400,000 * 9.8 * 50

Power = 1176000000000

Power = 1. 176 × 10^12 W

How many 15 W LED light bulbs could it power?

1176000000000 / 15 = 78400000000

= 78400000000 15 W bulbs

step by step explanation

simple distillation can be used when the temperature difference between the boiling points of two miscible liquid is at least 25°c. the temperature difference between the boiling points of kerosene and petrol is 25c. hence, this mixture can separated using simple distillation.

answer:

simple distillation

Answer:

Moles of hydrogen formed = 3.5 moles

Explanation:

Given that:-

Moles of aluminium= 4.0 mol

Moles of hydrogen bromide = 7.0 mol

According to the reaction:-

2 moles of aluminum react with 6 moles of hydrogen bromide

1 mole of aluminum react with 6/2 moles of hydrogen bromide

4 moles of aluminum react with (6/2)*4 moles of hydrogen bromide

Moles of hydrogen bromide = 12 moles

Available moles of hydrogen bromide = 7.0 moles

Limiting reagent is the one which is present in small amount. Thus, hydrogen bromide is limiting reagent. (7.0 < 12)

The formation of the product is governed by the limiting reagent. So,

6 moles of hydrogen bromide on reaction forms 3 moles of hydrogen

1 moles of hydrogen bromide on reaction forms 3/6 moles of hydrogen

7 moles of hydrogen bromide on reaction forms (3/6)*7 moles of hydrogen

Moles of hydrogen formed = 3.5 moles

Answer:

3.5 mol H2, HBr (limiting reactant)

Explanation:

4.0 mol Al × 3 mol H2/ 2 mol Al = 6.0 mol H2

7.0 mol HB ×3 mol H2/ 6mol HBr = 3.5 mol H2

Since 7.0mol of HBr will produce less H2 than 4.0mol of Al, HBr will be the limiting reactant, and the reaction will produce 3.5mol of H2.

Answer:

To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN

Explanation:

Dilution is the reduction of the concentration of a chemical in a solution and consists simply of adding more solvent.

In a dilution the amount of solute does not vary. But as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

In a solution it is fulfilled:

Ci* Vi = Cf* Vf

where:

• Ci: initial concentration
• Vi: initial volume
• Cf: final concentration
• Vf: final volume

In this case:

• Ci= 1.40 M
• Vi= 20 mL
• Cf= 0.088 M
• Vf= ?

Replacing:

1.40 M* 20 mL= 0.088 M* Vf

Solving:

Vf= 318.18 mL

To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN

Final answer:

To make a 0.0880 M solution of LiCN, you would need to dilute 20.0 mL of the 1.40 M solution to a final volume of 318.18 mL.

Explanation:

To dilute a solution, you can use the formula:

M1V1 = M2V2

where M1 and V1 are the initial molarity and volume, and M2 and V2 are the final molarity and volume. Rearranging the formula, we can solve for V2:

V2 = (M1 · V1) / M2

Plugging in the values given:

V2 = (1.40 M · 20.0 mL) / 0.0880 M = 318.18 mL

To make a 0.0880 M solution of LiCN, you would need to dilute 20.0 mL of the 1.40 M solution to a final volume of 318.18 mL.

Learn more about Dilution here:

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