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Veronika [31]
4 years ago
8

Could someone please help me with my chemistry hw please please please please please please please

Chemistry
1 answer:
IceJOKER [234]4 years ago
3 0
Jhgpqurhubfjnqjejfnbouqejdnfcnadvq f/g
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Which of the following elements does not form an ion with a charge of plus one hydrogen fluorine sodium or potassium
Reil [10]

Answer:

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6 0
3 years ago
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What is the equation for the acid dissociation constant, Ka, of HF?
VashaNatasha [74]

Answer:

hydrofluoric acid HF H+ + F- 7.2 × 10-4

4 0
3 years ago
A chemist has a 12.5 liter sample of neon gas at 1.00 atm and 30 c. If the chemist compresses the sample to 10.5 liters and hold
ycow [4]

Answer:

254.5 K

Explanation:

Data Given

initial volume V1 of neon gas = 12.5 L

final Volume V2 of neon gas = 10.5 L

initial Temperature T1 of neon gas = 30 °C

convert Temperature to Kelvin

T1 = °C +273

T1 = 30°C + 273 = 303 K

final Temperature T2 of neon gas = ?

Solution:

This problem will be solved by using Charles' law equation at constant pressure.

The formula used

                        V1 / T1 = V2 / T2

As we have to find out Temperature, so rearrange the above equation

                       T2 = V2 x T1 / V1

Put value from the data given

T2 = 10.5 L x 303 K / 12.5 L

T2 = 254.5K

So the final Temperature of neon gas = 254.5 K

8 0
3 years ago
The reaction a(g)⇌2b(g) has an equilibrium constant of k = 0.030. what is the equilibrium constant for the reaction b(g)⇌12a(g)?
aalyn [17]
First, we have to correct the equation in the question to b(g)⇆ 1/2 A(g)
at the first equation A(g)⇆ 2 B(g) so,
 Kc = [B]^2 [ A] = 0.03 

by reverse the equation 2B⇆ A 
∴ Kc(original) = [A] / [B]^2

        = 1/0.03 = 33 M^-1

and the new equation B⇆ (1/2) A 
So, the new Kc = √Kc(original = √33 
                ∴    KC  = 5.7
7 0
4 years ago
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One mole of ar initially at 305 k undergoes an adiabatic expansion against a pressure pexternal = 0 from a volume of 8.5 l to a
PilotLPTM [1.2K]
An adiabatic process is when the system is insulated that no heat is released to the surroundings. For this type of process, we have a derived formula written below:

(T₂/T₁)^C = (V₁/V₂)
where C = Cv/nR

From the complete problem shown in the attached picture, Cv = (3/2)R. Thus,
C= (3/2)/1 mol = 3/2

(T₂/305 K)^(3/2) = (8.5 L/82 L)
Solving for T₂,
<em>T₂ = 67.3 K</em>

3 0
3 years ago
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