How many moles of diphosphorus

Give a chemical formula for the compound formed when sulfur dioxide reacts with water

zavuch27 [327]

When sulfur dioxide reacts with water , the product is called sulphurous acid, H2SO3. The balanced equation is expressed in the following manner : SO2 + H2O = H2SO3. There is a one mole to one mole ratio between any of the reactants to product, sulphurous acid.

6 0

3 years ago

What type of energy does a person experience when jumping on a trampoline?

Fofino [41]

Answer:

Knetic

Explanation:

When you jump on a trampoline, your body has kinetic energy that changes over time. As you jump up and down, your kinetic energy increases and decreases with your velocity. Your kinetic energy is greatest, just before you hit the trampoline on the way down and when you leave the trampoline surface on the way up.

7 0

2 years ago

Wetlands provide habitats for many living things because of their sheltered waters and what other feature?

SIZIF [17.4K]

Answer:

large supply of nutrients.

Explanation:

In a wetland, the soil is covered by water or is almost covered by water. This water may be coming upwards from an underground aquifer. Wetlands are covered by water for most of the year.

They are sheltered waters and do provide habitats for many living things.

Nutrients such as; Carbon sulfur, phosphorus, carbon, and nitrogen are cycled within the soil of wetlands hence wetlands have a large supply of nutrients.

5 0

3 years ago

A sample of an unknown gas with a mass of 8.21 g has a volume of 4.8064 L when the temperature is 200oC and the pressure is 1.81

Pie

Answer:

The answer to your question is M = 36.49 g

Explanation:

Data

mass = 8.21 g

volume = 4.8064 L

Temperature = 200°C

Pressure = 1.816 atm

M = ?

Process

1.- Convert temperature to °K

°K = 273 + 200

°K = 473

2.- Calculate the number of moles

n = (PV)/RT

n = (1.816)(4.8064)/(0.082)(473)

n = 0.225

3.- Calculate the molar mass

M --------------- 1 mol

8.21 g ---------- 0.225 moles

M = (1 x 8.21)/0.225

M = 36.49 g

4 0

3 years ago

Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of

harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g) (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

$n = \frac{m}{M}$

Where:

m: is the mass

M: is the molar mass

For hydrogen we have:

$n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles$

And for nitrogen:

$n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles$

We can see in reaction (1) that 3 moles of hydrogen react with 1 mol of nitrogen, so the number of hydrogen moles needed to react nitrogen is:

$n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles$

Since we have 3.47 moles of hydrogen and we need 7.50 moles to react with all the mass of nitrogen, the limiting reactant is hydrogen.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that 3 moles of hydrogen produces 2 moles of ammonia, so:

$n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles$

Hence, hydrogen would produce 2.31 moles of ammonia.

Therefore, hydrogen is the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

I hope it helps you!

6 0

3 years ago