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Diano4ka-milaya [45]
2 years ago
15

Analyze How would tides be affected if the Moon was farther away from Earth?

Chemistry
2 answers:
strojnjashka [21]2 years ago
6 0

Answer: The farther away the moon is from earth the lower the tides would be.

Explanation: Because of the moons close proximity to Earth its gravity pulls up on water which causes the high and low tides, therefore if the moon is farther away its effect would be lessened.

trasher [3.6K]2 years ago
4 0

Answer:

Explanation:

Newton's law of gravity says that the attractive gravitational force is inversely proportional to the square of distance between two objects.

Earth tidal force is due to gravity pull from the Moon. That is why tides follow a monthly cycle according to the position of the Moon. If the Moon was farther away from Earth, tidal waves will be reduced.

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Answer:

each continent shifts 2-5 inches per year so once it was joined then it is what it is now.In a million years again it will comes closer and wjo knows what will happen.moutains will be formed as the tectonic plates will collide.

5 0
3 years ago
Water vapor turning to frost is an example of __________.
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Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
3 years ago
A solution of malonic acid, H2C3H2O4 , was standardized by titration with 0.1000 M NaOH solution. If 21.17 mL of the NaOH soluti
MrRa [10]

Answer: The molarity of the malonic acid solution is 0.08335 M

Explanation:

H_2C_3H_2O_4 +2NaOH\rightarrow Na_2C_3H_2O_4+2H_2O

To calculate the molarity of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2C_3H_2O_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=12.70mL\\n_2=1\\M_2=0.1000M\\V_2=21.17mL

Putting values in above equation, we get:

2\times M_1\times 12.70=1\times 0.1000\times 21.17\\\\M_1=0.08335M

Thus the molarity of the malonic acid solution is 0.08335 M

5 0
3 years ago
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