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Katen [24]
3 years ago
6

______is the condition of the atmosphere at any given time and place and ______is the weather over a long period of time (fill i

n the blanks using the terms weather, climate, or temperature.) ​
Chemistry
1 answer:
Nady [450]3 years ago
7 0

Answer:

Temperature is the condition of the atmosphere at any given time and place

and climate is the weather over a long period of time

Explanation:

I think so-

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In a normal cellular protein, where would you expect to find a hydrophobic amino acid like valine?
Irina-Kira [14]
In a branched chain of amino acids
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2CH3CHO+O2=> 2CH3COOH
Dovator [93]

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2CH3CHO + O2 → 2CH3COOH

7 0
2 years ago
How many moles of carbon are in 3.7 moles of C 8 H11 NO2
Dafna1 [17]
In every molecule of C_{8} H_{11}NO_2 there is 8 atoms of Carbon.

IF we have 3.7 moles of C_{8} H_{11}NO_2 to find the number of moles of Carbon, just multiply by 8

3.7 * 8 = 29.6 mol Carbon
6 0
3 years ago
Riley is scootering down the street before they come to a stop at a stop
11111nata11111 [884]

Answer:

just subrracted them and find out whats m

Explanation:

6 0
3 years ago
At equilibrium, the concentrations of the products and reactants for the reaction, H2 (g) + I2 (g)  2 HI (g), are [H2] = 0.106
lana [24]

Answer:

The new equilibrium concentration of HI: <u>[HI] = 3.589 M</u>          

Explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M        

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

<em>Given chemical reaction:</em> H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant (K_{c}) for the given chemical reaction, is given by the equation:

K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}

<u><em>At the original equilibrium state:</em></u>

K_{c} = \frac {(1.29\: M)^{2}}{(0.106\: M) \times (0.022\: M)}

K_{c} = \frac {1.6641}{0.002332} = 713.59

<u><em>Therefore, at the new equilibrium state:</em></u>

K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}

\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}

\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88

\Rightarrow [HI] = \sqrt {12.88} = 3.589 M

<u>Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M</u>

6 0
3 years ago
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