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damaskus [11]
3 years ago
12

It is winter, and you want to know where it might snow today. Look at the predicted temperature for each state. Then identify th

e states where it could snow today.
a
California, North Dakota, Washington

b
Michigan, California, Florida

c
North Dakota, Michigan

d
California, Florida, Washington
Chemistry
1 answer:
Alona [7]3 years ago
5 0

Answer:

reeeeeeeeeeeeeee

Explanation:

reeeeeeeeeeeeeeeeeeeeeee

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What was the original pressure in a rigid container if the new pressure is 82.5atm and the temperature was raised from 44C to 67
zhenek [66]

Answer: 54 atm

Explanation:

I did 67/82.5 then got 0.8121212121. I them divided 44 by 0.81212121 and got 54.1791044776

4 0
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ΔH for the formation of CuCl2 from its elements is -220.1 kJ/mol. How many kJ are associated with the formation of 0.30 mole of
otez555 [7]
0.3 * - 220.1 = - 66.03kJ
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3 years ago
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Electromagnetic waves heat earths surface through convection?
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Heat moves in three ways:Radiation, conduction, and convection. Radiation happens when heat moves as energy waves, call infrared waves,directly from its source to something else.This is how the heat from the Sun gets to Earth.In fact,all hot things radiate heat to cooler things:).hope this helps
4 0
4 years ago
Equal molar quantities of Ca2 and EDTA (H4Y) are added to make a 0.010 M solution of CaY2- at pH 10. The formation constant for
abruzzese [7]

Answer:

the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

Explanation:

Given the data in the question;

Ca^{2+ + y^{4- ⇄  CaY^{2-

Formation constant Kf

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) = 5.0 × 10¹⁰

Now,

[y^{4-] = \alpha _4CH_4Y; ∝₄ = 0.35

so the equilibrium is;

Ca^{2+ + H_4Y ⇄  CaY^{2- + 4H⁺

Given that; CH_4Y = Ca^{2+     { 1 mol Ca^{2+  reacts with 1 mol H_4Y  }

so at equilibrium, CH_4Y = Ca^{2+ = x

∴

Ca^{2+ + y^{4- ⇄  CaY^{2-

x        + x         0.010-x

since Kf is high, them x will be small so, 0.010-x is approximately 0.010

so;

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) =  CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ 0.010 / ( [x][ 0.35 × x] )  = 5.0 × 10¹⁰

⇒ 0.010 / 0.35x²  = 5.0 × 10¹⁰

⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 5.7142857 × 10⁻¹³

⇒ x = √(5.7142857 × 10⁻¹³)

⇒ x = 7.559 × 10⁻⁷

Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

8 0
3 years ago
Please help!!i NEED to get this right
Lana71 [14]
The correct answer is C
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2 years ago
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