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labwork [276]
3 years ago
11

What is the mass in grams of 2.0 moles of NO2

Chemistry
2 answers:
JulsSmile [24]3 years ago
8 0
Molar mass NO₂ = 46.0 g/mol

1 mole -------- 46.0 g
2.0 moles ----- ?

Mass (NO₂) = 2.0 x 46.0 / 1

=> 92.0 g

hope this helps!
blagie [28]3 years ago
7 0

Answer:

The mass of 2 moles of NO₂ is 92 grams.

Explanation:

Given that,

Number of moles, n = 2 moles

We need to find the mass of 2 moles of NO₂. We know that the number of moles is given by the formula as :

n=\dfrac{m}{M}

m is the given mass

M is the molar mass of NO₂, M=14+16\times 2=46\ g

m=n\times M\\\\m=2\times 46\\\\m=92\ g

So, the mass of 2 moles of NO₂ is 92 grams. Hence, this is the required solution.

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Molecular formula: C10H15Cl5
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"Calculate the number of millimoles in 500 mg of each of the following amino acids: alanine (MW = 89), leucine (131), tryptophan
Cloud [144]

Answer:

- Alanine =  5.61 mmoles

- Leucine = 3.81 mmoles

- Tryptophan = 2.45 mmoles

- Cysteine = 4.13 mmoles

- Glutamic acid = 3.40 mmoles

Explanation:

Mass / Molar mass = Moles

Milimoles = Mol . 1000

500 mg / 1000 = 0.5 g

- Alanine = 0.5 g / 89 g/m → 5.61x10⁻³ moles  . 1000 = 5.61mmoles

- Leucine = 0.5 g / 131 g/m → 3.81 x10⁻³ moles  . 1000 = 3.81 mmoles

- Tryptophan = 0.5 g / 204 g/m →  2.45x10⁻³ moles . 1000 = 2.45 mmoles

- Cysteine = 0.5 g / 121 g/m → 4.13x10⁻³ moles . 1000 = 4.13 mmoles

- Glutamic acid = 0.5 g 147 g/m → 3.40x10⁻³ moles . 1000 =  3.4 mmoles

5 0
3 years ago
As with other ionic compounds, potassium bromate, KBrO3, dissociates into ions when it dissolves in water. If 13.8 g of KBrO3 is
chubhunter [2.5K]

Answer:

ΔH of dissociation is 38,0 kJ/mol

Explanation:

The dissociation reaction of KBrO₃ is:

<em>KBrO₃ → K⁺ + BrO₃⁻ </em>

This dissolution consume heat that is evidenced with the decrease in water temperature.

The heat consumed is:

q = CΔTm

Where C is specific heat of water (4,186 J/mol°C)

ΔT is the temperature changing (18,0°C - 13,0°C = 5,0°C)

And m is mass of water (150,0 mL ≈ 150,0 g)

Replacing, heat consumed is:

q = 3139,5 J ≡ 3,14 kJ

13,8 g of KBrO₃ are:

13,8 g×(1mol/167g) = 0,0826 moles

Thus, ΔH of dissociation is:

3,14kJ / 0,0826mol = <em>38,0 kJ/mol</em>

<em></em>

I hope it helps!

3 0
3 years ago
Please answer the question
Tom [10]

Answer:

Stamens

Explanation:

7 0
3 years ago
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