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Georgia [21]
3 years ago
9

A sample of chlorine containing isotopes of mass numbers 35 and 37 was analysed in a mass-spectrometer. How many peaks correspon

ding to Cl+2 were recorded? A 2 B 3 C 4 D 5 Ans: B
Chemistry
1 answer:
Zolol [24]3 years ago
3 0
Three peaks corresponding to Cl+2 will be recorded. The peaks are for isotope 35, both 35 and 37 and for isotope 37. Mass spectrometer has the ability to detect and separate isotopes, even those differing by a single atomic mass unit. When chlorine isotopes are analysed by mass spectrometer, either peak M or M+2 can be obtained. The intensity ratio in the isotope pattern depends on the natural abundance of the isotopes.
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Please choose one of the choices
Georgia [21]

Answer:

A. the law of constant composition

Explanation:

The molecules in the container would have the same composition because they would have traded around atoms until an equilibrium was reached with every molecule having 1 Hydrogen and 1 Chlorine.

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Why does the moon's gravitational pull have more effect on Earth's tides than the sun's gravitational pull?
BaLLatris [955]
I believe it’s c because if you really read and close read it will make more sense
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Write the chemical equation for phosporic acid, h3po4, is produced through a chemical reaction between tetraphosporus decoxide a
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P_{4}O_{10} + 6H_{2}O ---\ \textgreater \  4H_{3}PO_{4}
6 0
3 years ago
What is the mass of 6.12 moles of arsenic (As)?
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Mole = Mass / Molar mass

6.12 moles = Mass / 74.92 g/mol

Mass = 6.12 moles x 74.92 g/mol

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3 years ago
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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.

6 0
4 years ago
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