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Alex73 [517]
3 years ago
11

Heat is being transferred from the ground to the air above it by conduction. Explain how this is possible.

Chemistry
2 answers:
tiny-mole [99]3 years ago
7 0
Hope this help enjoy ya day
Klio2033 [76]3 years ago
4 0

Answer:

Conduction directly affects air temperature only a few centimeters into the atmosphere. During the day, sunlight heats the ground, which in turn heats the air directly above it via conduction. At night, the ground cools and the heat flows from the warmer air directly above to the cooler ground via conduction.

Explanation:

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3 years ago
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How much of a 24-gram sample of Radium-226 will remain unchanged at the end of three half-life periods?
shutvik [7]

Answer:

The right answer is "3 g".

Explanation:

Given:

Initial mass substance,

M_0=24 \ g

By using the relation between half lives and amount of substances will be:

⇒ M=\frac{M_0}{2^n}

        =\frac{24}{2^3}

        =3 \ g

Thus, the above is the correct answer.

8 0
3 years ago
Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find
FromTheMoon [43]

<u>Answer:</u> The concentration of SO_4^{2-} at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

       H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)

            0.025          0.025       0.025

Equation for the second dissociation of sulfuric acid:

                    HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)

<u>Initial:</u>            0.025            0.025      

<u>At eqllm:</u>      0.025-x          0.025+x        x

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

We know that:

Ka_2\text{ for }H_2SO_4=0.01

Putting values in above equation, we get:

0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of SO_4^{2-} at equilibrium is 0.00608 M

4 0
3 years ago
Convert each into decimal form.<br>a) 1.56× 10^3<br>b) 0.56×10-4​
Lera25 [3.4K]

Answer: A = 1560

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Explanation: brainlest please

3 0
3 years ago
How many σ and π bonds are present in a molecule of cumulene?
dimaraw [331]

A hydrocarbon with three or more consecutive (cumulative) double bonds is known as a cumulene. They are analogous to allenes, only exhibiting a more elongated chain. The basic molecule in this category is butatriene, which is also simply known as cumulene.  

In the structure of a cumulene, there are 3 double bonds and 4 single bonds. The double bond comprises 1 sigma bond, and 1 pi bond and 4 hydrogen bond produces a sigma bond with carbon. Thus, the molecule of cumulene comprises 7 sigma bonds and 3 pi bonds.  


3 0
3 years ago
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