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2 years ago

Calculate the volume in L of 12.9 g of NO2 at STP.

Chemistry

1 answer:

d1i1m1o1n [39]2 years ago

3 0

Answer:

6.28 L

Explanation:

Given data:

Volume of NO₂ = ?

Mass of NO₂ = 12.9 g

Temperature and pressure = ?

Solution:

First of all we will calculate the number of moles of gas.

Number of moles = mass/molar mass

Number of moles = 12.9 g/ 46 g/mol

Number of moles = 0.28 mol

At standard temperature and pressure 1 mole of gas occupy 22.4 L volume.

Thus,

0.28 mol × 22.4 L/1 mol

6.27 L ≅ 6.28 L

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How many particles are in a 151 g sample of Li2O?

neonofarm [45]

Answer:

3.052 × 10^24 particles

Explanation:

To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)

The mass of Li2O given in this question is as follows: 151grams.

To convert this mass value to moles, we use;

moles = mass/molar mass

Molar mass of Li2O = 6.9(2) + 16

= 13.8 + 16

= 29.8g/mol

Mole = 151/29.8g

mole = 5.07moles

number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23

= 30.52 × 10^23

= 3.052 × 10^24 particles.

4 0

2 years ago

23.9 km + 15 km

Ugo [173]

Answer:

the original answer is 38.9km (3sf)

4 0

3 years ago

If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe

pishuonlain [190]

Answer:

Explanation:

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.

Mᵣ: 30.01 32.00 46.01

2NO + O₂ ⟶ 2NO₂

Mass/g: 80.00 16.00

2. Calculate the moles of each reactant

$\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}$

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

$\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}$

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

$\text{Moles of NO}_{2} = \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

$\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}$

(b) Mass of NO reacted

$\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}$

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0

2 years ago

Beryllium oxide, Beo, is an electrical insulator. How

egoroff_w [7]

Answer:

There are 10.0 moles of beryllium oxide in a 250 grams sample of the compound.

Explanation:

We can calculate the number of moles (η) of BeO as follows:

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

Where:

m: is the mass = 250 g

M: is the molar mass = 25.0116 g/mol

Hence, the number of moles is:

$\eta = \frac{250 g}{25.0116 g/mol} = 10.0 moles$

Therefore, there are 10.0 moles of beryllium oxide in a 250 grams sample of the compound.

I hope it helps you!

3 0

2 years ago

The descriptions below explain two ways that water is used by plants on a sunny day. I. In a process called transpiration, some

eduard

The correct answer for this question is this one: "<span>In transpiration, because some of its properties change, water undergoes a physical change but keeps its identity. In photosynthesis, because its identity changes, water undergoes a chemical change</span>. "
Hope this helps answer your question and have a nice day ahead.

8 0

2 years ago

Read 2 more answers