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nevsk [136]
3 years ago
14

What is needed for a wind farm to be effective

Chemistry
1 answer:
qwelly [4]3 years ago
7 0

Answer:

for a effective wind farm you would need wind turbines a windy area a rotation powered generator and something to use it on

Explanation:

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Assume that you have no limit on the amount of money you can spend on a tennis racket. What kind
Taya2010 [7]

Answer:

I would buy a high quality racket and a racket that would enhance your abilities more. I would buy one that helps you in any way possible if you just buy one for looks then ur screwed

Explanation:

7 0
3 years ago
Which of the following are correct about nodal planes in molecular orbitals? There may have more than one correct answers. Pictu
azamat

Explanation:

I'll give you a real answer later need points for test questions sorry

4 0
4 years ago
What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11
lutik1710 [3]
The first dissociation for H2X:
                        H2X +H2O ↔ HX + H3O
initial                0.15                     0      0
change             -X                     +X      +X
at equlibrium 0.15-X                  X        X
because Ka1 is small we can assume neglect x in H2X concentration
     Ka1      = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15) 
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
        HX + H2O↔ X^2 + H3O
    8.2x10^-4          Y         8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2       = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m


4 0
4 years ago
Explain how venus rotation is different from earths
saul85 [17]
It has the slowest rotation of all planets. 

like 240 earth days...in retrograde clockwise turns
7 0
3 years ago
Read 2 more answers
I NEED HELP PLEASE, THANKS! :)
marin [14]

Answer:

\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}

Explanation:

1. Calculate the moles of copper(II) hydroxide

\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}

2. Calculate the molecules of copper(II) hydroxide

\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}

6 0
3 years ago
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