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fredd [130]
3 years ago
8

Formed by a spinning column of air that touches the ground in 1% of thunderstorms

Chemistry
2 answers:
dlinn [17]3 years ago
8 0
I believe the answer is C. Tornado
LUCKY_DIMON [66]3 years ago
6 0
C. tornado is your answer
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leva [86]

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0.054 moles

Explanation:

It is the rounded off answer.

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Which of the following factors affect the vapor pressure of a liquid? Check all that apply. View Available Hint(s) Check all tha
trasher [3.6K]

Answer:

Vapour pressure of liquid is affected by its TEMPERATURE, TYPE OF LIQUID, ADDITION OF SOLUTES.

Explanation:

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid. Mathematically, the vapor pressure of liquid is written as:

In (Pvap) = -ΔHvap / R * (1/T) + C

ΔHvap is the enthalpy of vaporization and it is constant for specific liquid, T is the temperature and C is a constant. Temperature therefore affect the vapor pressure of a liquid as increase in temperature can cause an increase in vapor pressure.

The type of liquid also affect the vapor pressure. If the molecules bind to each other strongly in the liquid, it increases vapor pressure to be exerted by the liquid but if the type of liquid is one with weak forces of attraction, the vapor pressure is reduced.

Addition of solute to the liquid increases the boiling temperature of liquids thereby reducing it's vapor pressure.

6 0
3 years ago
Given the equilibrium constants for the following two reactions in aqueous solution at 25 ∘C HNO2(aq)H2SO3(aq)⇌⇌H+(aq) + NO2−(aq
krok68 [10]

Answer : The value of K_c for the final reaction is, 184.09

Explanation :

The equilibrium reactions in aqueous solution are :

(1) HNO_2(aq)\rightleftharpoons H^+(aq)+NO_2^-(aq)         K_{c_1}=4.5\times 10^{-4}

(2) H_2SO_3(aq)\rightleftharpoons 2H^+(aq)+SO_3^{2-}(aq)         K_{c_2}=1.1\times 10^{-9}

The final equilibrium reaction is :

2HNO_2(aq)+SO_3^{2-}(aq)\rightleftharpoons H_2SO_3(aq)+2NO_2^-(aq)         K_{c}=?

Now we have to calculate the value of K_c for the final reaction.

Now equation 1 is multiply by 2 and reverse the equation 2, we get the value of final equilibrium reaction and the expression of final equilibrium constant is:

K_c=\frac{(K_{c_1})^2}{K_{c_2}}

Now put all the given values in this expression, we get :

K_c=\frac{(4.5\times 10^{-4})^2}{1.1\times 10^{-9}}=184.09

Therefore, the value of K_c for the final reaction is, 184.09

4 0
3 years ago
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