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Alona [7]
3 years ago
9

Gases in the atmosphere that trap solar energy

Chemistry
2 answers:
Korolek [52]3 years ago
7 0

Answer:

d

Explanation:

A greenhouse gas (sometimes abbreviated GHG) is a gas that absorbs and emits radiant energy within the thermal infrared range. Greenhouse gases cause the greenhouse effect on planets. The primary greenhouse gases in Earth's atmosphere are water vapor (H. 2O), carbon dioxide (CO. 2), methane (CH.

nalin [4]3 years ago
6 0
D greenhouse gases is a gas that is trapped in the solar energy
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garri49 [273]

b is the correct answer

6 0
3 years ago
If one mole of na3po4·3h2o is heated extensively, how many moles of water are released?
Aleks04 [339]

When one mole of Na3PO4.3H2O is heated extensively, three moles of water are released.

The water molecules in Na3PO4.3H2O are called molecules of water of crystallization. These molecules are not covalently bonded to the Na3PO4 molecule. They are only loosely attached to the substance.

Strong heating will drive away these molecules of water of crystallization to give three moles of water in the product.

Hence, when one mole of Na3PO4.3H2O is heated extensively, three moles of water are released.

Learn more: brainly.com/question/14252791

4 0
2 years ago
How many atoms are in 4H3BO3
maksim [4K]

1 mole of h3bo3...........6.023*10²³ each h and B and 0 so we will have 3hydrogen+ 1 B+3 oxygen = 7*6.023*10²³ atoms

1 mole .......7*6.023*10²³atoms

4 moles ........x atoms

x=4*7*6.023*10²³.

3 0
3 years ago
4. Solve the following heat flow problem, being sure to show all your work (you may either type your
Viktor [21]

Answer:

0.70 J/g.°C

Explanation:

Step 1: Given data

  • Mass of graphite (m): 402 g
  • Heat absorbed (Q): 1136 J
  • Initial temperature: 26°C
  • Final temperature: 30 °C
  • Specific heat of graphite (c): ?

Step 2: Calculate the specific heat of graphite

We will use the following expression.

Q = c × m × ΔT

c = Q / m × ΔT

c = 1136 J / 402 g × (30°C - 26°C)

c = 0.70 J/g.°C

5 0
3 years ago
Water (with density of 1000 kg/m3) with the mass flowrate of 10 kg/sec is flowing into an empty tank. The outlet volumetric flow
Montano1993 [528]

Explanation:

Apply the mass of balance as follows.

   Rate of accumulation of water within the tank = rate of mass of water entering the tank - rate of mass of water releasing from the tank

         \frac{d}{dt}(\rho V) = 10 - \rho \times (0.01 h)

      \rho A_{c} \frac{dh}{dt} = 10 - (0.01) \rho h

   \frac{dh}{dt} + \frac{0.01 \rho h}{\rho A_{c}} = \frac{10}{\rho A_{c}}

          [/tex]\frac{dh}{dt} + \frac{0.01}{0.01}h[/tex] = \frac{10}{\rho A_{c}}

                       A_{c} = 0.01 m^{2}

              \frac{dh}{dt} + h = 1

                  \frac{dh}{dt} = 1 - h

               \frac{dh}{1 - h} = dt  

                \frac{ln(1 - h)}{-1} = t + C      

Given at t = 0 and V = 0  

                         A \times h = 0  

 or,                     h = 0

                 -ln(1 - h) = t + C

Initial condition is -ln(1) = 0 + C

                                C = 0  

                So,   -ln(1 - h) = t

or,                      t = ln (\frac{1}{1 - h})  ........... (1)

(a)    Using equation (1) calculate time to fill the tank up to 0.6 meter from the bottom as follows.

                    t = ln (\frac{1}{1 - h})  

                     t = ln (\frac{1}{1 - 0.6})  

                        = ln (\frac{1}{0.4})

                        = 0.916 seconds

(b)   As maximum height of water level in the tank is achieved at steady state that is, t = \infty.  

                    1 - h = exp (-t)

                    1 - h = 0  

                         h = 1

Hence, we can conclude that the tank cannot be filled up to 2 meters as maximum height achieved is 1 meter.

                 

8 0
3 years ago
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