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timurjin [86]
3 years ago
14

The average temperature at the South Pole In January is - 35.4 °C.

Chemistry
1 answer:
IrinaVladis [17]3 years ago
5 0

Answer:

-31.72°F

Explanation:

(-35.4°C × 9/5) + 32 = -31.72°F

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What is the frequency of a photon of light (in Hz) that has an energy of 2.50 x 10-18 J?
ohaa [14]

Answer:

The answer is

3.77 \times  {10}^{15}  \:  \:  \: Hz

Explanation:

The frequency of the photon can be found by using the formula

f =  \frac{E}{h}  \\

where

E is the energy

f is the frequency

h is the Planck's constant which is

6.626 × 10-³⁴ Js

From the question

E = 2.50 × 10^-18 J

So we have

f =  \frac{2.50 \times  {10}^{ - 18} }{6.626 \times  {10}^{ - 34} }  \\

We have the final answer as

3.77 \times  {10}^{15}  \:  \:  \: Hz

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Answer:

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Explanation:

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What animal did Darwin study in the Galapagos Islands?
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Answer:

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Explanation:

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Calcium carbonate decomposes to form calcium oxide and carbon dioxide, like this:
Leni [432]

Answer: The value of the equilibrium constant Kc for this reaction is 0.088

Explanation:

Molarity=\frac{x}{M\times V_s}

where,

x = given mass

M = molar mass

V_s = volume of solution in L

Equilibrium concentration of CaCO_3 = \frac{25.3}{100\times 9.0}=0.028M

Equilibrium concentration of CaO = \frac{14.9}{56\times 9.0}=0.029M

Equilibrium concentration of CO_2 = \frac{33.7}{44\times 9.0}=0.085M

The given balanced equilibrium reaction is,

CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CaO]\times [CO_2]}{[CaCO_3]}  

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

K_c=\frac{0.029\times 0.085}{0.028}=0.088

7 0
3 years ago
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