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1 year ago

In the laboratory you dissolve 22.0 g of nickel(II) nitrate in a volumetric flask and add water to a total volume of 125

Chemistry

1 answer:

Ahat [919]1 year ago

3 0

Answer:

<u>1.364 M</u>.

Explanation:

Molarity formula: M= n/v, where n is moles of solute, and v is liters of solution.

Now, we need to convert the grams of nickel to moles and the volume of water to liters.

125mL/1000= 0.125 L.

To convert nickel grams to moles, we need to take a look at it's chemical formula, which is:

$Ni(NO_{3} )_{2}$

Now we count how many molecules of each element we have:

Ni= 1

N= 2

O= 6

Calculate the weight (g) of each element (the values of g/mol can be found on the periodic table and they may vary slightly between one table and the other):

Ni: (1) (58.6934)= 58.6934

N= (2) (14.007)= 28.014

O= (6) (15.999)= 95.994

Sum all the values to obtain the total weight of 1 mole of this compound:

58.6934+28.014+95.994= 129(g/mole)

Now that we know the that 129 grams equal 1 mole of nickel(II) nitrate, we can convert the 22.0 g to moles:

129g ------- 1 mole

22.0g ----- x

x= (22*1)/129= 0.1705 moles.

Now, we have all the values needed to calculate the molarity of this solution. All we have to do is substitute the values in the formula:

M= (0.1705 moles) / (0.125 L)= <u>1.364 M</u>.

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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

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3 years ago

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Juliette [100K]

Answer:

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2 years ago

A student is asked to identify and element that is pale yellow brittle solid and does not conduct electricity. At which location

disa [49]

Answer:

Far right

Explanation:

The qualities indicated here are of the element Sulphur (S) and any element of same qualities i.e color pale yellow, nature solid, electrical conductivity none, should be placed at far right corner of the periodic table or position 3 mentioned in the attached picture.

Hope it help!