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Artist 52 [7]
3 years ago
11

The high temperatures on venus are caused by the _______.

Chemistry
2 answers:
raketka [301]3 years ago
7 0

As sunlight heats Venus' surface, the surface radiates infrared energy that is kept from escaping the planet by dense carbon dioxide atmosphere. This is called the greenhouse effect.

guajiro [1.7K]3 years ago
4 0

The High tempatures on Venus are caused by the Greenhouse effect

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What is the percentage of water in the hydrate cocl2 • 6h2o
lord [1]

Answer:

52.17%

Explanation:

COCl2.6H20

C=12,O=16,Cl=35.5,H=1

Relative molecular mass of COCl2.6H2O= 12+16+71+6(2+16) = 99 + 108= 207g

Relative molecular mass of 6H2O = 108g

Percentage of water = (108/207 )*100

= 52.17%

8 0
3 years ago
Which statement about subatomic particles is true? (2 points) Protons are the only subatomic particles to have charge. Electrons
Alona [7]
Protons are not the only subatomic particle that has a charge, electrons do to and they are negatively charged. Neutrons are situated in the nucleus of the atom, which means that they do not orbit it. Subatomic particles have different masses. 

You are then left with Electrons are the subatomic particles with the smallest mass as your answer. 
6 0
3 years ago
N2 (g) + 2O2(g) = 2NO2 (g) ΔH = 66.4 kJ 2NO (g) + O2 (g) = 2NO2 (g) ΔH = -114.2 kJ the enthalpy of the reaction of the nitrogen
RSB [31]

Answer:

ΔH  = 180.6 kJ

Explanation:

Given that:

N2 (g) + 2O2(g) = 2NO2 (g)           ΔH = 66.4 kJ

<u>2NO (g) + O2 (g) = 2NO2 (g)         ΔH = -114.2 kJ                     </u>

N2 (g) + O2 (g) = 2NO (g)              ΔH  = ????

The subtraction of both equations would yield the unknown ΔH , therefore:

ΔH = 66.4 - ( - 114.2 kJ)

ΔH  = 180.6 kJ

3 0
3 years ago
Read 2 more answers
What are the two processes underlying this image that turn peat into coal
Travka [436]

Answer:

Peatification and coalification

Explanation:

https://energyeducation.ca/encyclopedia/Coal_formation

8 0
4 years ago
Read 2 more answers
Calculate the solubility of mn(oh)2 in grams per liter when buffered at ph=7.0. assume that buffer capacity is not exhausted
Vanyuwa [196]
When PH + POH = 14 
∴ POH = 14 -7 = 7

when POH = -㏒[OH-]

          7    = -㏒ [OH-]
∴[OH-] = 10^-7

by using ICE table:

           Mn(OH)2(s) ⇄  Mn2+ (aq)  + 2OH-(aq)
initial                              0                     10^-7
change                           +X                      +2X
Equ                                 X                  (10^-7 + 2X)

when Ksp = [Mn2+][OH-]^2

when Ksp of Mn(OH)2 = 4.6 x 10^-14

by substitution:

4.6 x 10^-14 = X*(10^-7+2X)^2  by solving this equation for X

∴ X =2.3 x 10-5 M

∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol

= 0.002 g/ L
3 0
4 years ago
Read 2 more answers
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