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2 years ago

How many moles of N are in 0.223 g of N2O

1 answer:

4 0

Answer:
There are twice the molar quantity of nitrogen atoms in nitrous oxide, i.e.
1×10−2⋅mol
Moles of nitrous oxide

=

0.217
⋅
g
44.013
⋅
g
⋅
m
o
l
−
1

=

5
×
10
−
3
⋅
m
o
l
.
Given the composition of
N
2
O
, there are this
2
×
5
×
10
−
3
⋅
m
o
l
of nitrogen atoms.

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Example of false solutions and true solutions

Kamila [148]

Answer:

answer in expl.

Explanation:

example of false solution.- Earth is not square

true eg. - we live in earth

3 0

3 years ago

A lithium flame has a characteristic red color due to emissions of wavelength 671 nm. What is the mass equivalence of 1 mol of p

e-lub [12.9K]

Answer:

1.98x10⁻¹² kg

Explanation:

The energy of a photon is given by:

E= hc/λ

h is Planck's constant, 6.626x10⁻³⁴ J·s

c is the speed of light, 3x10⁸ m/s

and λ is the wavelenght, 671 nm (or 6.71x10⁻⁷m)

E = 6.626x10⁻³⁴ J·s * 3x10⁸ m/s ÷ 6.71x10⁻⁷m = 2.96x10⁻¹⁹ J

Now we multiply that value by Avogadro's number, to calculate the energy of 1 mol of such protons:

1 mol = 6.023x10²³ photons

2.96x10⁻¹⁹ J * 6.023x10²³ = 1.78x10⁵ J

Finally we calculate the mass equivalence using the equation:

E=m*c²

m=E/c²

m = 1.78x10⁵ J / (3x10⁸ m/s)² = 1.98x10⁻¹² kg

6 0

3 years ago

Which of the following ions is formed when a base is dissolved in a solution?

Aneli [31]

Answer:

$\huge\boxed{\sf OH-}$

Explanation:

According to Arrhenius concept of acid and base:

"When a base in a solution, produces/yields OH- (Hydroxide) ions."

So, when a base is dissolved in a solution, it produces OH- ions.

For example:

NaOH ⇄ Na⁺ + OH⁻ (So, it is a base)

$\rule[225]{225}{2}$

Hope this helped!

3 0

2 years ago

Substitute natural gas (SNG) is a gaseous mixture containing CH4(g) that can be used as a fuel. One reaction for the production

Lynna [10]

Answer:

ΔH° of the reaction is -747.54kJ

Explanation:

Based on gas law, it is possible to find the ΔH of a reaction using ΔH of half reactions.

Using the reactions:

(1) C(graphite) + 1/2O₂(g) → CO(g) ΔH° = -110.5 kJ

(2) CO(g) + 1/2O₂(g) → CO₂(g) ΔH° = -283.0 kJ

(3) H₂(g) + 1/2O₂(g) → H₂O(l) ΔH° = -285.8 kJ

(4) C(graphite) + 2H₂(g) → CH₄(g) ΔH° = -74.81 kJ

(5) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH° = -890.3 kJ

The sum of 4×(4) + (5) gives:

4C(graphite) + 8H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -74.81 kJ ×4 - 890.3 kJ = -1189.54kJ

Now, this reaction - 4×(1) gives:

4CO(g) + 8H₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -1189.54kJ - 4×-110.5 = -747.54kJ

Thus ΔH° of the reaction is -747.54kJ

3 0

2 years ago

The empirical mass is a whole number multiple of the molar mass true or false?

elena-14-01-66 [18.8K]

Answer:

True

Explanation:

6 0

2 years ago