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irga5000 [103]
3 years ago
8

Which is greater, the moons period of rotation or its period of revolution?

Chemistry
2 answers:
Alexxx [7]3 years ago
4 0

Answer:

The answer is C. They are equal

balu736 [363]3 years ago
3 0
C. they are equal

This is because it takes about the same amount of time for the moon to rotate around its axis that it does to rotate around the Earth.
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Ammonia (NH3) reacts with sulfuric acid to form ammonium sulfate. How many grams of ammonium sulfate are obtained from 8.73 g of
Gwar [14]
I hope you understand my working:
1) Finding the mol of NH3 to find the mol of (NH4)2SO4 (ammonium sulfate)
2) Mr of (NH4)2SO4
3) Theoretical yield: The actual grams of (NH4)2SO4 produced when reacting 0.514 mol of NH3 to 0.514 mol H2SO4
4) Using formula of (given grams)/(theoretical grams or actual grams) * 100 = 73%
5) Basic algebra

4 0
3 years ago
Wich stament describes to organ systems working together to get rid of waste played by cells
IceJOKER [234]

Answer:

C. Kidneys filter wastes from the bloodstream and produce urine

Explanation:

5 0
2 years ago
Read 2 more answers
Classify each of the following as a pure substance or a mixture. If a mixture, indicate whether it is homogeneous or heterogeneo
Mashutka [201]

Answer:

a) Heterogeneous mixture (b) Homogenous mixture (c) Pure substance (d) Pure substance

Explanation:

Homogenous mixtures contains mixture of substances with similar proportions while Heterogenous mixture contains substances with a varying proportion.

8 0
3 years ago
What is the ph of a buffer prepared by adding 0.607 mol of the weak acid ha to 0.609 mol of naa in 2.00 l of solution? the disso
Paraphin [41]

Given:

0.607 mol of the weak acid

0.609 naa

2.00 liters of solution

 

The solution for finding the ph of a buffer:

[HA] = 0.607 / 2.00 = 0.3035 M 
[A-]= 0.609/ 2.00 = 0.3045 M 
pKa = 6.25 

pH = 6.25 + log 0.3045/ 0.3035 = 6.25 is the ph buffer prepared.

6 0
3 years ago
Calculate Ho298 for the process
Inga [223]

Explanation:

As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

Sb + \frac{3}{2}Cl_2 \rightarrow SbCl_{3}    \Delta H^0_1 =  -314 kJ  ..........(1)

SbCl_{3} + Cl_2 \rightarrow SbCl_{5}    \Delta H^0_2 = -80kJ   ..............(2)

The final reaction is as follows:  

Sb + \frac{5}{2}Cl_{2} \rightarrow SbCl_{5}  \Delta H^0_3 = ?  .............(3)

Therefore, adding (1) and (2) we get the final equation (3) and value of \Delta H^{0}_{3} at 298 K will be as follows.

             \Delta H^{0}_{3} = \Delta H^{0}_{1} + \Delta H^{0}_{2}    

                       = -314 kJ + (-80) kJ

                       = -394 kJ

Thus, we can conclude that H^{o} at 298 K for the given process is -394 kJ.

4 0
3 years ago
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