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2 years ago

How many moles of nitrogen, n, are in 83.0 g of nitrous oxide, n2o?

1 answer:

7 0

The moles of nitrogen n in 83.0 g of N2O is calculated as below

find the moles of N2O
moles = mass/molar mass

= 83 g/44g/mol= 1.886 moles
the moles of Nitrogen in N2O is therefore

=1.886 moles x2(since there are two atom of N in N2O) =3.772 moles

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A grocery store has 75 dozen packages of hotdogs In stock. If there are 10 hot dogs in each package, how many moles of hot dogs

Viktor [21]

you can just divide 75 and 10

6 0

2 years ago

Read 2 more answers

Calculate the number of atoms in 2.5 mol manganese.

Leokris [45]

Answer:

1.5 × 10²⁴ atoms Mn

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

2.5 mol Mn

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

$2.5 \ mol \ Mn(\frac{6.022 \cdot 10^{23} \ atoms \ Mn}{1 \ mol \ Mn} )$ = 1.5055 × 10²⁴ atoms Mn

Step 4: Check

We are given 2 sig figs. Follow sig fig rules and round.

1.5055 × 10²⁴ atoms Mn ≈ 1.5 × 10²⁴ atoms Mn

6 0

1 year ago

At low temperatures and high pressures ethene gas C2H4(g), does not behave like an ideal gas. Use chemical principles to explain

Luba_88 [7]

Answer:

The particles of an ideal gas have no volume and no attractions for each other. In a real gas, however, the molecules do have a measurable volume. The molecules of real cases have intermolecular attractions for each other.

An ideal gas behaves like a real gas under the conditions of low temperature and high pressure.

This is because at low temperature and high pressure molecules of gas will have negligible kinetic energy and strong force of attraction.

Thus ethene gas does not behave like an ideal gas at low temperatures and high pressures.

6 0

2 years ago

Anne prepares a solution of copper nitrate with a concentration of 30 g/dm³. What mass of copper nitrate did she add to 100cm³ o

strojnjashka [21]

Answer:

3 g

Explanation:

Concentration of copper nitrate needed = 30 g/dm³

Volume of water = 100 cm³ = 0·1 dm³

( ∵ 1 dm³ = 1000 cm³)

Let the mass of copper nitrate that should be added to 100 cm³ water be m g

m ÷ 0·1 = 30

∴ m = 3 g

∴ Mass of copper nitrate that should be added to 100 cm³ of water = 3 g

7 0

2 years ago

Estimate ΔG°rxn for the following reaction at 449.0 K. CH2O(g) + 2 H2(g) → CH4(g) + H2O(g) ΔH°= -94.9 kJ; ΔS°= -224.2 J/K

Sliva [168]

This equation relates all three variables, so just plug in all of the given values. Don’t forget to convert deltaS into kJ/K because Gibbs free energy is measured in kJ. The K used in temperature cancels out the K in kJ/K, leaving only kJ as your units. Don’t forget sig figs!

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2 years ago