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4 years ago

10

To draw the Lewis structure of the polyatomic ion,CIO3 you would have to _____ those in the structures of Cl, O, O, and O. Selec

t one:
a. add a proton to
b. subtract one electron from
c. subtract a proton from
d. add one electron to

1 answer:

NikAS [45]4 years ago

3 0

D. Add one electron to,

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How much faster does ethane (C2H6) gas travel compared to chlorine gas?

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Answer:

$\large \boxed{\text{54 $\, \%$ faster }}$

Explanation:

$v_{\text{rms}} \propto \sqrt{\dfrac{3RT}{M}$

if temperature is constant.

$v_{\text{rms}} \propto \sqrt{\dfrac{1}{M}$

if we are comparing two gases,

$\dfrac{v_{2}}{v_{1}} = \sqrt{\dfrac{M_{1}}{M_{2}}}$

Let chlorine be Gas 1 and ethane be Gas 2

Data:

M₁ = 70.91 g/mol

M₂ = 30.07 g/mol

Calculation

$\begin{array}{rcl}\dfrac{v_{2}}{v_{1}} & = & \sqrt{\dfrac{M_{1}}{M_{2}}}\\\\& = & \sqrt{\dfrac{70.91}{30.07}}\\\\& = & \sqrt{2.358}\\\\& = & \mathbf{1.54}\\\end{array}\\\text{Ethane molecules travel at 1.54 times the speed of chlorine molecules}\\\text{or $\large \boxed{\textbf{54 $\%$ faster }}$ than chlorine molecules}$

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Answer:

Your questions requires diagrams of the cell to get which one is on the left or right. However, see the attached file below

The correct answer is (d) the left half-cell will decrease in concentration; and the right half-cell will increase in concentration.

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The concentration of the Pb2+ increases in the oxidation half cell while the concentration of the Pb2+ decreases in the reduction half cell during the reaction.

In the Left Beaker (Left half cell), their is less concentration

Pb(s) ---> Pb2+(aq) + 2 e- Concentration of Pb2+(aq) increase ; Electrons going out from this side

In the Right Beaker (right half cell), their is more concentration

Pb2+(aq) + 2 e- ---> Pb(s) Concentration of Pb2+(aq) decrease ; Electrons coming in to this side

Electrons will flow from Left to Right direction.

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