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3 years ago

What is the mass in grams of one nitrogen molecule?

Chemistry

1 answer:

Oksana_A [137]3 years ago

6 0

Answer:

The mass of one Nitrogen molecule is option C

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At the ph found in cells (about 7.0) what happens to the amino group on an amino acid

Andrew [12]

This means that the amino acid is neutral due to the pH equals 7

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3 years ago

At constant temperature, if a gas occupies 312 mL at a pressure of 1.60 atm, what pressure is necessary for this gas to occupy a

inn [45]

Answer:

P₂ = 1.0 atm

Explanation:

Boyles Law problem => P ∝ 1/V at constant temperature (T).

Empirical equation

P ∝ 1/V => P = k(1/V) => k = P·V => for comparing two different case conditions, k₁ = k₂ => P₁V₁ = P₂V₂

Given

P₁ = 1.6 atm

V₁ = 312 ml

P₂ = ?

V₂ = 500 ml

P₁V₁ = P₂V₂ => P₂ = P₁V₁/V₂ =1.6 atm x 312 ml / 500ml = 1.0 atm

5 0

3 years ago

16. Mass = 10g

inessss [21]

Actual volume=Final Volume-initial volume

$\\ \sf\longmapsto 50ml-30ml=20ml$

Now

$\\ \sf\longmapsto Density=\dfrac{Mass}{Volume}$

$\\ \sf\longmapsto Density=\dfrac{10}{20}$

$\\ \sf\longmapsto Density=2g/ml$

3 0

3 years ago

2. force → the rate at which an object changes position; distance divided by time

yaroslaw [1]

Answer:

2. True

3. True

4. False

Explanation:

4 0

3 years ago

A 57.0 mL sample of a 0.120 M potassium sulfate solution is mixed with 35.5 mL of a 0.118 M lead(II) acetate solution and the fo

Katarina [22]

Answer:

Limiting reagent = lead(II) acetate

Theoretical yield = 1.2704 g

% yield = 78.09 %

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium sulfate :

Molarity = 0.120 M

Volume = 57.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 57.0×10⁻³ L

Thus, moles of potassium sulfate:

$Moles=0.120 M \times {57.0\times 10^{-3}}\ moles$

Moles of potassium sulfate = 0.00684 moles

For lead(II) acetate :

Molarity = 0.118 M

Volume = 35.5 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 35.5×10⁻³ L

Thus, moles of lead(II) acetate :

$Moles=0.118 \times {35.5\times 10^{-3}}\ moles$

Moles of lead(II) acetate = 0.004189 moles

According to the given reaction:

$K_2SO_4_{(aq)}+Pb(C_2H_3O_2)_2_{(aq)}\rightarrow 2KC_2H_3O_2_{(s)}+PbSO_4_{(aq)}$

1 mole of potassium sulfate react with 1 mole of lead(II) acetate

0.00684 moles potassium sulfate react with 0.00684 mole of lead(II) acetate

Moles of lead(II) acetate = 0.004189 moles

Limiting reagent is the one which is present in small amount. Thus, lead(II) acetate is limiting reagent. ( 0.004189 < 0.00684)

The formation of the product is governed by the limiting reagent. So,

1 mole of lead(II) acetate gives 1 mole of lead(II) sulfate

0.004189 mole of lead(II) acetate gives 0.004189 mole of lead(II) sulfate

Molar mass of lead(II) sulfate = 303.26 g/mol

Mass of lead(II) sulfate = Moles × Molar mass = 0.004189 × 303.26 g = 1.2704 g

Theoretical yield = 1.2704 g

Given experimental yield = 0.992 g

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (0.992/1.2704 g) × 100 = 78.09 %</u>

3 0

3 years ago