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Firdavs [7]
2 years ago
10

What mass of nickel (Ni) is in a 2.4 Kg sample of propanol if the concentration is 20 ppb ? (atomic mass of Ni = 58.69)

Chemistry
1 answer:
asambeis [7]2 years ago
8 0

Answer:

The mass of nickel is 48μg

Explanation:

Parts per billion is a way to describe small concentrations and is defined as the ratio between μg of solute and kg of solvent.

If a solution of nickel in propanol is 20ppb, contains 20μg of nickel in 1 kg of propanol.

Thus, a sample of 2.4kg of propanol will contain:

2.4kg × (20μg nickel / 1kg) = 48μg nickel

<h3>The mass of nickel is 48μg</h3>
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Jill then wrote the formulas for all the possible compounds that sodium can form with the other ions. What formulas did she writ
sp2606 [1]

Explanation:

Sodium has atomic number of 11 and its electronic configuration is given by:

[Na]=1s^22s^22p^63s^1

The nearest stable electronic configuration to sodium is of the neon. So, in order to attain stability of noble gas it will loose its single electron.

Na\rightarrow Na^++1e^-

[Na^+]=1s^22s^22p^63s^0

Sodium has single valency that is 1.

Let nbe the valency of the ion 'X'

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

Na_nX_1

So, the formulas for all the possible compounds that sodium can form with the other ions will be:

Na_nX_1

4 0
3 years ago
What are the correct coefficients when this equation is balanced?
larisa86 [58]

4, 3, 1

Explanation:

Sb has four on the right, so it needs four on the left. It's all alone, so <u>4</u>.

O2 comes in pairs, so you only need <u>3</u> of those to get six oxygens.

And 1 is the place holder for the right side since we got the left to match perfectly.

6 0
2 years ago
I added and multiplied i still don't know
hjlf

Answer:

it'a answer number 2

Explanation: you divide the mass by volume and 32.2 divided by 4 is 8.05

8 0
2 years ago
How does a filter separate mixtures like sand and water?​
Talja [164]

Answer:

Filtration is a method for separating an insoluble solid from a liquid. When a mixture of sand and water is filtered: the sand stays behind in the filter paper (it becomes the residue ) the water passes through the filter paper (it becomes the filtrate )

Explanation:

8 0
2 years ago
A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so
Xelga [282]

Answer : The equilibrium concentration of SCN^- in the trial solution is 4.58\times 10^{-8}M

Explanation :

First we have to calculate the initial moles of Fe^{3+} and SCN^-.

\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}

\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol

and,

\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}

\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol

The given balanced chemical reaction is,

Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)

Since 1 mole of Fe^{3+} reacts with 1 mole of SCN^- to give 1 mole of FeSCN^{2+}

The limiting reagent is, SCN^-

So, the number of moles of FeSCN^{2+} = 0.0020 mmole

Now we have to calculate the concentration of FeSCN^{2+}.

\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M

Using Beer-Lambert's law :

A=\epsilon \times C\times l

where,

A = absorbance of solution

C = concentration of solution

l = path length

\epsilon = molar absorptivity coefficient

\epsilon and l are same for stock solution and dilute solution. So,

\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}

For trial solution:

The equilibrium concentration of SCN^- is,

[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]

[SCN^-]_{initial} = 0.00050 M

Now calculate the [FeSCN^{2+}].

C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M

Now calculate the concentration of SCN^-.

[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]

[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)

[SCN^-]_{eqm}=4.58\times 10^{-8}M

Therefore, the equilibrium concentration of SCN^- in the trial solution is 4.58\times 10^{-8}M

5 0
3 years ago
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