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Reptile [31]
3 years ago
15

When the ground becomes saturated in a rain storm and the excess cannot be soaked up, the water will travel along the ground unt

il it hits a low-lying area or a larger body of water. The water that is not absorbed into the ground, including the larger body of water, is _______.
Chemistry
2 answers:
dalvyx [7]3 years ago
8 0

The answer is surface water

OLEGan [10]3 years ago
7 0

Answer:

The answer is surface water

Explanation:

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If a straight-chain hydrocarbon is a gas at room temperature, how many carbon atoms will it have?
Orlov [11]

The fewer the carbon atoms, the closer it is  to being a gas. The only one you have to check out is A which is hexane. You know that gasoline at the pumps has 8 carbons and its a liquid. So B and C are both not gases  because they are above 8.

C6 (hexane) is a liquid at room temperature not a gas.

The answer is D. If there is a gas present, it must be C3

6 0
3 years ago
Which of the following is NOT how compounds are formed?
cricket20 [7]
Compounds are not formed by accepting electrons. Number 1 and 3 are correct.
5 0
3 years ago
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What is the change in enthalpy when 4.00 mol of sulfur trioxide decomposes to sulfur dioxide and oxygen gas? 250 2(g) + O2(g) ®
lidiya [134]

Answer: A. -396 kJ

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

2SO_2(g)+O_2 (g)\rightarrow 2SO_3(g)  \Delta H^0=198kJ

Reversing the reaction, changes the sign of \Delta H

2SO_3(g)\rightarrow 2SO_2(g)+O_2 (g)  \Delta H^0=-198kJ

On multiplying the reaction by 2, enthalpy gets multiplied by 2:

4SO_3(g)\rightarrow 4SO_2(g)+2O_2 (g)   \Delta H=2\times -198kJ=-396kJ

Thus the enthalpy change for the reaction 4SO_3(g)\rightarrow 4SO_2(g)+2O_2 (g)  is -396 kJ.

8 0
3 years ago
Which tool would best allow you to make observations of an object's color and texture?
strojnjashka [21]

Answer: I would say it is the D

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For a reaction to be spontaneous under standard conditions at all temperatures, the signs of δh° and δs° must be __________ and
Shalnov [3]
For a reaction to be spontaneous under standard conditions at all temperatures, the signs of ΔH° and ΔS° must be negative and positive, respectively.
<span>Gibbs free energy (G) determines if reaction will proceed spontaneously, if </span>ΔG is negative, reaction is spontaneous.<span>
ΔG = ΔH - T·ΔS.
ΔG - changes in Gibbs free energy.
ΔH - changes in enthalpy.
ΔS - changes in entropy.
T is temperature in Kelvins.</span>

8 0
3 years ago
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