what modifications to either or both of the half-cell reactions would kim need to make if she wanted to use them to produce equa
1 answer:
We have to add the both half cell equations and eliminate the number of electrons lost/gained.
<h3>What modification must Kim make to the equations?</h3>The term redox reaction is a type of reaction that occurs when an electron is lost or gained in a reaction system. We can see that in this reaction, zinc looses two electron which are gained by copper.
If we want to obtain the equation 4.9 which is the overall equation of the redox reaction from the various half cell equations then we have to add the both half cell equations and eliminate the number of electrons lost/gained.
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Answer: 27.09 ppm and 0.003 %.
First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.
Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.
So, according to the <em>law of ideal gases,</em>
PV = nRT
where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)
The moles of CO will be,
n = 35 mg x x
→ n = 0.00125 mol
We clear V from the equation and substitute P = 0.92 atm and
T = -30 ° C + 273.15 K = 243.15 K
V =
→ V = 0.0271 L
As 1000 cm³ = 1 L then,
V = 0.0271 L x = 27.09 cm³
<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>
c = 27 cm³ / m³ = 27 ppm
<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:
c = 27.09 x
x 100%
c = 0.003 %
So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.
To get the number of gold atoms, you have to divide the mass of the gold by the mass of the gold atom. It follows this simple equation 
.
Let x be the number of gold atoms. Plug in the values to a calculator.
x =
Both have the same units so the unit gram(g) can be cancelled.
x then would be equal to 1.53x10^22. So there are 1.53x10^22 atoms of gold in 5 g of gold
Let x be the number of gold atoms. Plug in the values to a calculator.
x =
Both have the same units so the unit gram(g) can be cancelled.
x then would be equal to 1.53x10^22. So there are 1.53x10^22 atoms of gold in 5 g of gold