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anyanavicka [17]
3 years ago
12

What is the difference between mass number and atomic mass (sometimes also called atomic weight)? Boron has an atomic mass of 10

.811. What does this mean?
Chemistry
1 answer:
RUDIKE [14]3 years ago
3 0

Answer:

<h3> I think....atomic mass: it is the average of total masses of isotopes in the element..... mass no: it is the total no.of protons and neutrons in atomic nuclei of abundant isotope in the element</h3>

Explanation:

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Blast furnaces extra pure iron from the Iron(IIl)oxide in iron ore in a two step sequence. In the first step, carbon and oxygen
OLga [1]

Answer:

5.9 kg  

Explanation:

We must work backwards from the second step to work out the mass of oxygen.

1. Second step

Mᵣ:                                     55.84

            Fe₂O₃ + 3CO  ⟶  2Fe  +  3CO₂

m/kg:                                    7.0

(a) Moles of Fe

\text{Moles of FeO} = \text{7000 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.84 g Fe}} = \text{125 mol Fe}

(b) Moles of CO

\text{Moles of CO} = \text{125 mol Fe} \times \dfrac{\text{3 mol CO}}{\text{2 mol Fe}} = \text{188 mol CO}

However, this is the theoretical yield.

The actual yield is 72. %.

We need more CO and Fe₂O₃ to get the theoretical yield of Fe.

(c) Percent yield

\begin{array}{rcl}\text{Percent yield} &=& \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \, \%\\\\ 72. \, \% & = & \dfrac{\text{188 mol}}{\text{actual yield}} \times 100 \,\%\\\\0.72 &= &\dfrac{\text{188 mol}}{\text{actual yield}}\\\\\text{Actual yield} & = & \dfrac{\text{188 mol}}{0.72}\\& = & \textbf{261 mol}\\\\\end{array}

We must use 261 mol of CO to get 7.0 kg of Fe.

2. First step

Mᵣ:                32.00

            2C   +  O₂   ⟶  2CO

n/mol:                             261

(a) Moles of O₂

\text{Moles of O}_{2} = \text{261 mol CO} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol CO}} = \text{131 mol O}_{2}

(b) Mass of O₂

\text{Mass of O}_{2}= \text{131 mol O }_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol  O}_{2}} = \text{4180 g O}_{2}

However, this is the theoretical yield.

The actual yield is 71. %.

We need more C and O₂ to get the theoretical yield of CO.

(c) Percent yield

\begin{array}{rcl}71. \, \% & = & \dfrac{\text{188 mol}}{\text{actual yield}} \times 100 \,\%\\\\0.71 &= &\dfrac{\text{4180 g}}{\text{actual yield}}\\\\\text{Actual yield} & = & \dfrac{\text{4180 g}}{0.71}\\\\& = & \text{5900 g}\\& = & \textbf{5.9 kg}\\\end{array}

We need 5.9 kg of O₂ to produce 7.0 kg of Fe.

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Gold forms a substitutional solid solution with silver. Calculate the number of gold atoms per cubic centimeter (in atoms/cm3) f
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Answer:

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Silver: 4.8 x 10²² atoms/cm³

Explanation:

100 g of the alloy will have 29 g of Au and 71 g of Ag.

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29 g Au ______  x

x = 1.5 cm³

10.49 g Ag ____ 1 cm³

71 g Ag _______   y

y = 6.8 cm³

The total volume of 100g of the alloy is x+y = 8.3 cm³.

Gold:

196.97 g Au____ 6.022 x 10²³ atoms Au

29 g Au _______ w

w = 8.9 x 10²² atoms Au

8.9 x 10²² atoms Au ____ 8.3 cm³

           A                    ____ 1 cm³

A = 1.1 x 10²² atoms Au

Silver:

107.87 g Ag____ 6.022 x 10²³ atoms Ag

71 g Ag _______ w

w = 4.0 x 10²³ atoms Ag

4.0 x 10²³ atoms Ag ____ 8.3 cm³

          B                    ____ 1 cm³

B = 4.8 x 10²² atoms Ag

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Answer:True

Explanation:

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