Answer: 21 lawns per week
Explanation:
The average weekly rate refers to how many lawns were mowed per week given that 233 were done in 11 weeks.
Rate will be given by;
= Lawns mowed / Weeks taken
= 233 / 11
= 21 lawns per week
Answer:
yes
Explanation:
Kids like fun things, and school is not fun , so if kids got to go to school on a zipline, attendance would be through the roof!
Partial pressure of gas A is 1.31 atm and that of gas B is 0.44 atm.
The partial pressure of a gas in a mixture can be calculated as
Pi = Xi x P
Where Pi is the partial pressure; Xi is mole fraction and P is the total pressure of the mixture.
Therefore we have Pa = Xa x P and Pb = Xb x P
Let us find Xa and Xb
Χa = mol a/ total moles = 2.50/(2.50+0.85) = 2.50/3.35 = 0.746
Xb = mol b/total moles = 0.85/(2.50+0.85) = 0.85/3.35 = 0.254
Total pressure P is given as 1.75 atm
Pa = Xa x P = 0.746 x 1.75 = 1.31atm
Partial pressure of gas A is 1.31 atm
Pb = Xb x P = 0.254 x 1.75 = 0.44atm
Partial pressure of gas B is 0.44 atm.
Learn more about Partial pressure here:
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<span>You have to use a Newman projection to make sure that the H on C#2 is anti-coplanar with the Br on C#1. (Those are the two things that are going to be eliminated to make the alkene.)
My Newman projection looks like this when it's in the right configuration:
Front carbon (C#2) has ethyl group straight up, H down/left, and CH3 down/right
Back carbon (C#1) has H straight down, Ph up/left, and Br up/right.
Then when you eliminate the H from C#2 and the Br from C#1, you will have Ph and the ethyl group on the same side of the molecule, and you'll have the remaining H and CH3 on the same side of the molecule.
This is going to give you (Z)-2-methyl-1-phenyl-1-butene.</span>
Answer:
the concentration of bicarbonate is <em>[HCO₃⁻] = 0,03996 M </em>and carbonate is <em>[CO₃²⁻] = 3,56x10⁻⁵ M.</em>
Explanation:
Carbonate-bicarbonate is:
HCO₃⁻ ⇄ CO₃²⁻ + H⁺ With pka = 10,25
Using Henderson-Hasselbalach formula:
pH = pka + log₁₀![\frac{[CO_{3}^{2-}]}{[HCO_{3}^-]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO_%7B3%7D%5E%7B2-%7D%5D%7D%7B%5BHCO_%7B3%7D%5E-%5D%7D)
7,2 = 10,25 + log₁₀![\frac{[CO_{3}^{2-}]}{[HCO_{3}^-]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO_%7B3%7D%5E%7B2-%7D%5D%7D%7B%5BHCO_%7B3%7D%5E-%5D%7D)
8,91x10⁻⁴ =
<em>(1)</em>
Also:
0,040 M = [CO₃²⁻] + [HCO₃⁻] <em>(2)</em>
Replacing (2) in 1:
<em>[HCO₃⁻] = 0,03996 M</em>
Thus:
<em>[CO₃²⁻] = 3,56x10⁻⁵ M</em>
I hope it helps.