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zysi [14]
3 years ago
15

A certain ore is 20.7% nickel by mass. How many kilograms of this ore would you need to dig up to have 70.0 g of nickel?

Chemistry
1 answer:
LenKa [72]3 years ago
6 0

Answer:

0.3382 kg

Explanation:

From the illustration:

<em>100g of the ore would contain 20.7 g of nickel by mass</em>. In order to obtain 70.0 g of nickel, the amount of the ore needed would be:

          100 x 70.0/20.7 = 338.16 g

According to conversion units:

<em>1 g = 0.001 kg</em>

Hence,

338.16 g = 338.16 x 0.001 = 0.3382 kg

<em>Therefore, </em><em>0.3382 kg</em><em> of the ore would be required in order to obtain 70.0 g of nickel.</em>

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The isotope of carbon used in archaeological dating is 14^6C . How many protons, neutrons, and electrons does an atom of 14^6C h
vampirchik [111]

Answer:

6

8

6

Explanation:

 Isotope given:

                       ¹⁴₆C

In specie written as this;  

   Superscript  = Mass number

   Subscript  = Atomic number

To find the protons, it is the same as the atomic number;

                Protons  = Atomic number  = 6

Neutrons have no charges;

 Neutrons  = Mass number - Atomic number  =

  Neutrons  = 14 - 6  = 8

The number of electrons is the same as the atomic number = 6

8 0
3 years ago
How many ug of nickel (Ni) are required to make 25.00 nanoliters of a 1.25 mol/L solution? Be sure to report your answer to the
devlian [24]

volume of Ni = 25 nL = 25 x 10⁻⁹ L

mol Ni = 25 x 10⁻⁹ L x 1.25 mol/L = 3.125 x 10⁻⁸

mass = mol x Ar Ni

mass = 3.125 x 10⁻⁸ x 59 g/mol

mass = 1.84 x 10⁻⁶ g = 1.84 μg

4 0
2 years ago
Read 2 more answers
How demanding is lime and cement in our world's economy?
Serggg [28]
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6 0
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A reaction that occurs in the internal combustion engine is n2(g) + o2(g) ⇌ 2 no(g) (a) calculate δh o and δs o for the reaction
jekas [21]
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ΔrH = 2 mol · 90.3 kJ/mol - (0 kJ/mol + 0 kJ/mol).
ΔrH = 180.6 kJ.
2) ΔS = 2mol·ΔS(NO) - (ΔS(O₂) + ΔS(N₂)).
ΔS = 2mol · 210.65 J/mol·K - (1mol · 205 J/mol·K + 1 mol · 191.5 J/K·mol).
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2 years ago
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vovikov84 [41]

Answer:

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3

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Hope this helped!

4 0
2 years ago
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