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Yuliya22 [10]
2 years ago
13

Wind power is:

Chemistry
2 answers:
12345 [234]2 years ago
7 0
Answer) All of the above
Margarita [4]2 years ago
6 0
I think the answer is a
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How many molecules are in 23 Moles of oxygen o2
loris [4]

Answer:

1.38×10^25 molecules

Explanation:

Applying n= (no. of molcules)/NA

23 = N/6.02×10^23

= 1.38×10^25 molecules

6 0
3 years ago
This answer please :))
r-ruslan [8.4K]

Answer: ya this one

Explanation: this is the one

4 0
3 years ago
The equilibrium constant for the chemical equation N2(g) + 3H2(g) ⇌ 2NH3(g) and Kp=0.174 at 243°C. Calculate the value of Kc for
Mashcka [7]

Answer:

The Kc of this reaction is 311.97

Explanation:

Step 1: Data given

Kp = 0.174

Temperature = 243 °C

Step 2: The balanced equation

N2(g) + 3H2(g) ⇌ 2NH3(g)

Step 3: Calculate Kc

Kp = Kc *(RT)^Δn

⇒ with Kp = 0.174

⇒ with Kc = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 Latm/Kmol

⇒ with T = the temperature = 243 °C = 516 K

⇒ with Δn = number of moles products - moles reactants  2 – (1 + 3) = -2

0.174 = Kc (0.08206*516)^-2

Kc = 311.97

The Kc of this reaction is 311.97

3 0
3 years ago
Water's ability to dissolve a wide variety of molecules is important, but more important is the hydrophobic effect, which drives
Sliva [168]

Answer:

d. Hydrophobic molecules are attracted to each other.

Explanation:

The term “hydrophobic effect” is associated with the spontaneous tendency of macromolecules, such as proteins, to prefer a conformation in an aqueous medium, with hydrophobic groups facing the interior of the mac romolecule, favoring attractive intramolecular interactions, and hydrophilic groups exposed on the surface, for maximize interactions with water molecules in the medium. This is because the hydrophobic molecules are attracted to each other, allowing them to turn inward.

8 0
3 years ago
A 5.00 L sample of air at 0 C is warmed to 100.0 C. What is the new volume of the air? First, identify V1.
Paladinen [302]

Answer : The new volume of the air is, 6.83 L

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=5.00L\\T_1=0^oC=(0+273)K=273K\\V_2=?\\T_2=100^oC=(100+273)K=373K

Putting values in above equation, we get:

\frac{5.00L}{273K}=\frac{V_2}{373K}\\\\V_2=6.83L

Therefore, the new volume of the air is, 6.83 L

6 0
3 years ago
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