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poizon [28]
3 years ago
15

What happens to thermal energy from the sun as a result of the greenhouse effect?

Chemistry
1 answer:
labwork [276]3 years ago
8 0
The temperature rises


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Two stones resembling diamonds are suspected of being fakes. To determine if the stones might be real, the mass and volume of ea
Charra [1.4K]

Answer:

Stone A

Explanation:

Measuring density is an easiest way to determine if two similar looking substances are same or not. Here also we need to perform the density test for each stone that is suspected to be fake diamond. We will calculate the density of each stone and compare it with the density of original diamond.

Density is calculated using the formula

Density=\frac{Mass}{Volume}

It has been given in the question that both the substances have same volume of 0.15 cm^3.

Density of stone A = \frac{0.52}{0.15} = 3.47 gcm^-^3 or after rounding off we get 3.5 gcm^-^3

Density of stone B = \frac{0.42}{0.15} = 2.8 gcm^-^3

It is clear from the above calculation that the stone A has same density as the diamond but stone B lacks behind in density.

So, stone A could be the real diamond.

4 0
3 years ago
The bouncing of light when is changes media is called what
USPshnik [31]
IT will be Refraction of light
3 0
2 years ago
Read 2 more answers
Consider the reaction. Upper H subscript 2 upper o (g) plus upper C l subscript 2 upper O (g) double-headed arrow 2 upper H uppe
dedylja [7]

Answer:

Equilibrium constant for the reaction is 0.0892

Explanation:

This is the reaction of equilibrium

         H₂O      +     Cl₂O     ⇄    2HClO

Eq   0.077M       0.077M           0.023M

Let's make the expression for Kc

Kc = [HClO]² / [Cl₂O] . [H₂O]

Kc= 0.023² / 0.077 . 0.077 = 0.0892

5 0
3 years ago
Read 2 more answers
At the normal melting point of NaCl, 801 degrees C, its enthalpy of fusion is 28.8 kJ / mol. The density of the solid is 2.165 g
melisa1 [442]

<u>Answer:</u> The increase in pressure is 0.003 atm

<u>Explanation:</u>

To calculate the final pressure, we use the Clausius-Clayperon equation, which is:

\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

P_1 = initial pressure which is the pressure at normal boiling point = 1 atm

P_2 = final pressure = ?

\Delta H = Enthalpy change of the reaction = 28.8 kJ/mol = 28800 J/mol     (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 801^oC=[801+273]K=1074K

T_2 = final temperature = (801+1.00)^oC=802.00=[802+273]K=1075K

Putting values in above equation, we get:

\ln(\frac{P_2}{1})=\frac{28800J/mol}{8.314J/mol.K}[\frac{1}{1074}-\frac{1}{1075}]\\\\\ln P_2=3\times 10^{-3}atm\\\\P_2=e^{3\times 10^{-3}}=1.003atm

Change in pressure = P_2-P_1=1.003-1.00=0.003atm

Hence, the increase in pressure is 0.003 atm

6 0
3 years ago
What are members one through six of the alkyne family
Luda [366]
Hydrocarbon hydrogen thats all ik sorry
4 0
3 years ago
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