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cestrela7 [59]
3 years ago
13

1.In the outline find your section on soil horizons. In it you will find a diagram representing a view of a soil cut away. Which

of the layers in this example represents the horizon that is alive with roots, fungi, bacteria, worms, small arthropods like spiders and so on?
A. O
B. A
C. B
D. C

2.In the outline find your section on soil horizons. In it you will find a diagram representing a view of a soil cut away. Which of the layers in this example represents the horizon that is very thin and contains large amounts of organic matter?
A. O
B.A
C.B
D.C


3.True or false: In the outline find your section on soil horizons. In it you will find a diagram representing a view of a soil cut away. Horizon B in this example would be very difficult for soil organisms to move through because it would be a very hard layer.
A. True
B.False
Chemistry
1 answer:
gogolik [260]3 years ago
3 0
What grade level 
 is this?
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How are transuranium elements made?
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Explanation:

The transuranium elements are produced by the capture of neutrons

<u>Hope</u><u> </u><u>it</u><u> </u><u>will</u><u> </u><u>help</u><u> </u><u>you</u>

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Use the given data at 500 K to calculate ΔG°for the reaction
Anton [14]

Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

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What is true about ionic compounds?
bezimeni [28]
It’s D I am pretty sure.
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1. Some athletes have as little as 3.0% body fat. If such a person has a body mass of 65 kg, how many pounds of body fat does th
77julia77 [94]
1. 3.0% ----> 3.0 kg fat= 100 kg body weigh
also remember that 1 kg= 2.20 lbs

65 kg  \frac{3.0 kg}{100 kg} x  \frac{2.20 lb}{1 kg} = 4.29 lbs

2. 0.94 g/mL----> 0.94 grams= 1 mL
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1kg= 1000 grams

3 Liters  \frac{1000 mL}{1 L} x   \frac{0.94 grams}{1 mL} x  \frac{1 kg}{1000 g} x   \frac{2.20 lbs}{1 kg}  = 6.20 lbs
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