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2 years ago

What is relative isotopic mass

Chemistry

1 answer:

lidiya [134]2 years ago

8 0

Answer:

Relative atomic mass or atomic weight is a dimensionless physical quantity defined as the ratio of the average mass of atoms of a chemical element in a given sample to the atomic mass constant. The atomic mass constant is defined as being 1/12 of the mass of a carbon-12 atom.

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GUYS PLEASE HELP IM CRYNG

erica [24]

Answer:

A.) 8.796 B.) 234780 C.) 25.8

Explanation:

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5 0

2 years ago

How many moles of NH3 would be formed from the complete reaction of 16.0 g H2?

natima [27]

Taking into account the reaction stoichiometry, 5.33 moles of NH₃ are formed from the complete reaction of 16 grams of H₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

N₂ + 3 H₂ → 2 NH₃

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

N₂: 1 mole
H₂: 3 moles
NH₃: 2 moles

The molar mass of the compounds is:

N₂: 14 g/mole
H₂: 2 g/mole
NH₃: 17 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

N₂: 1 mole ×14 g/mole= 14 grams
H₂: 3 moles ×2 g/mole= 6 grams
NH₃: 2 moles ×17 g/mole=34 grams

<h3>Mass of NH₃ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 6 grams of H₂ form 2 moles of NH₃, 16 grams of H₂ form how many moles of NH₃?

$moles of NH_{3}= \frac{16 grams of H_{2} x2moles of NH_{3}}{6 grams of H_{2}}$

<u><em>moles of NH₃= 5.33 moles</em></u>

Then, 5.33 moles of NH₃ are formed from the complete reaction of 16 grams of H₂.

Learn more about the reaction stoichiometry:

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4 0

1 year ago

Which statement is true? Postively charged objects attract other positively charged objects negatively charged objects attract o

ANEK [815]

Answer:

Explanation:one is true because if it is positiveit would not be attracting it would be sepperating

8 0

2 years ago

Which of the following compounds contains the Lead (II) ion?<br> PbO<br> PbCl4<br> Pb2O<br> Pb2S

KATRIN_1 [288]

<span>PbO Let's look at each of the 4 compounds and see what's needed. PbO. * Oxygen has a valance shell that's missing 2 electrons and wants to get those 2 elections. Lead donates them, so you have a Lead (II) ions. This is a correct choice. PbCl4 * Chlorine wants to grab 1 electron to fill it's valance shell and Lead donates that election. However, there's 4 chlorine atoms and every one of them wants and electron, and lead is donating all 4 of the desired electrons making the Lead (IV) ion. So this is a bad choice. Pb2O * Oxygen still wants 2 electrons and gets them from the lead. But there's 2 lead atoms and each of them donates 1 election making for 2 Lead(I) ions. So this too is a bad choice. Pb2S * Sulfur is in the same column of the periodic table as oxygen and if this compound were to exist would have similar properties as Pb2O and would have Lead(I) ions. So this is a bad choice.</span>

6 0

3 years ago

Read 2 more answers

1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0

2 years ago