What 4 elements have the same number of energy levels

The fertilizer ammonium sulfate, (NH4)2SO4, is prepared by the reaction between ammonia (NH3) and sulfuric acid:

kumpel [21]

The mass of ammonia required to produce 2.40 × 10⁵ kg of (NH₄)₂SO₄ is 6.18 * 10⁴ Kg of ammonia.

<h3>What mass in kilograms of ammonia are required to produce 2.40 × 10⁵ kg of (NH₄)₂SO₄?</h3>

The mass of ammonia required to produce 2.40 × 10⁵ kg of (NH₄)₂SO₄ is determined from the mole ratio of the reaction.

The mole ratio of the reaction is obtained from the balanced equation of the reaction given below:

2NH₃(g) + H₂SO₄(aq) → (NH₄)₂SO₄(aq)

Mole ratio of NH₃ and (NH₄)₂SO₄ is 2: 1

Mass of 2 moles of ammonia = 2 * 17 = 34 g

Mass of 1 mole of (NH₄)₂SO₄ = 132 g

Mass of ammonia required = 34/132 * 2.40 × 10⁵ kg

Mass of ammonia required = 6.18 * 10⁴ Kg of ammonia.

In conclusion, the mole ratio is used to determine the mass of ammonia required.

Learn more about mole ratio at: brainly.com/question/19099163

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4 0

2 years ago

How to find molecular weight

Nastasia [14]

Answer:

Using the periodic table of the elements to find atomic weights, we find that hydrogen has an atomic weight of 1, and oxygen's is 16. In order to calculate the molecular weight of one water molecule, we add the contributions from each atom; that is, 2(1) + 1(16) = 18 grams/mole.

Explanation:

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3 0

3 years ago

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq) , as described b

Nostrana [21]

Answer:

0.676 grams of manganese (IV) oxide should be added.

Explanation:

Moles of chlorine gas = n

Volume of the chlorine gas = V = 205 mL = 0.205 L

Pressure of the chlorine gas = 705 Torr = $\frac{705}{760}atm=0.928 atm$

1 atm = 760 Torr

Temperature of the chlorine gas = T = 25°C = 25 + 273 K = 298 K

$PV=nRT$ ( ideal gs equation)

$n=\frac{PV}{RT}=\frac{0.928 atm\times 0.205 L}{0.0821 atm L/mol K\times 298 K}=0.00777 mol$

$MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+2H_2O(l)+Cl_2(g)$

According to reaction, 1 mole of chlorine gas is obtained from 1 mole of manganese(IV) oxide,then 0.00777 moles of chlorine gas will be obtained from :

$\frac{1}{1}\times 0.00777 mol=0.00777 mol$ of manganese (IV) oxide

Mass of 0.00777 moles of manganese (IV) oxide:

0.00777 mol × 87 g/mol = 0.676 g

0.676 grams of manganese (IV) oxide should be added.

3 0

3 years ago

The rock in a particular iron ore deposit contains 82 % Fe2O3 by mass. How many kilograms of the rock must be processed to obtai

Zinaida [17]

Answer:

Mass of rock = 4533.4 kg

Explanation:

Percent composition is percentage by the mass of element present in the compound.

Given that the $Fe_2O_3$ is 82 % by mass.

It means that 82 kg of $Fe_2O_3$ is present in 100 kg of rock

To find the mass of rock which contains 2600 kg of iron

2600 kg = 2600*1000 g

2 moles of iron are present in 1 mole of $Fe_2O_3$

Molar mass of iron = 55.845 g/mol

Mass of 2 moles = 55.845 * 2 = 111.69 g/mol

Molar mass of $Fe_2O_3$ = 159.69 g/mol

It means,

111.69 g of iron are present in 159.69 g of $Fe_2O_3$

1 g of iron is present in 159.69/111.69 g of $Fe_2O_3$

2600*1000 g is present in (159.69/111.69)*2600*1000 g of $Fe_2O_3$

Mass of $Fe_2O_3$ = 3717378.47 g =3717.38 kg

Thus,

82 kg of $Fe_2O_3$ is present in 100 kg of rock

1 kg of $Fe_2O_3$ is present in 100/82 kg of rock

3717.38 kg of $Fe_2O_3$ is present in (100/82)*3717.38 kg of rock

Mass of rock = 4533.4 kg

5 0

3 years ago

Read 2 more answers

What is the concentration of nitrate ion in a 425 mL solution containing 32.0 g of Mg(NO3)2 (M = 148.3)?

Brums [2.3K]

<span>There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. Calculations are as follows:

Molarity of NO3 ion = 32.0 g Mg(NO3)2 ( 1 mol / 148.3 g ) ( 2 mol NO3 / 1 mol Mg(NO3)2) / .425 L
MOlarity of NO3 ion = 1.02 M </span><span>
</span>

8 0

3 years ago

Read 2 more answers