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coldgirl [10]
2 years ago
10

Please someone help! I think I did the wrong subject...but I REALLYY need help on this! I'LL MAKE THIS 40 POINTS!!! HERE ARE THE

QUESTIONS!
If the mass of the sun is 1x, at least one planet will fall into the habitable zone if I place a planet in orbits___, ____, ____, and ____, and all planets will orbit the sun successfully.

If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits___, ____, ____, and ____, and all planets will orbit the sun successfully.

If the mass of the sun is 3x, at least one planet will fall into the habitable zone if I place a planet in orbits___, ____, ____, and ____, and all planets will orbit the sun successfully.



PLEASE PLEASE HELP!
Chemistry
2 answers:
Nikitich [7]2 years ago
8 0

Answer:

Explanation:

adelina 88 [10]2 years ago
4 0

Answer:

give the other person brainliest

Explanation:

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In North America, where is population density the highest ?
zepelin [54]

Near the coasts and Great Lakes.

3 0
3 years ago
As stated in the article, “As Sticky as a Gecko . . . but Ten Times Stronger!,” the adhesive the researchers developed sticks be
alexandr1967 [171]

Answer:

shear adhesion

Explanation:

edge

6 0
3 years ago
Read 2 more answers
Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)
Dovator [93]

Answer:

1. 0.97 V

2. Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)

Explanation:

In this case, we can start with the <u>half-reactions</u>:

Ag^+~_(_a_q_)->~Ag_(_s_)

Al_(_s_)~->~Al^+^3~_(_a_q_)

With this in mind we can <u>add the electrons</u>:

Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)  <u>Reduction</u>

Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~ <u>Oxidation</u>

The reduction potential values for each half-reaction are:

Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_) - 0.69 V

Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_) -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:

Al_(_s_)~->~Al^+^3~+~2e^-~ +1.66 V

Finally, to calculate the overall potential we have to <u>add</u> the two values:

1.66 V - 0.69 V = <u>0.97 V</u>

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)

I hope it helps!

3 0
3 years ago
Given that 25.0 mL of mercury has a mass of 340.0 g, calculate (a) the density of mercury and (b) the mass of 120.0 mL of mercur
natita [175]

Answer :

(a) The density of mercury is, 13.6 g/ml

(b) The mass of 120.0 ml of mercury is, 1632 grams

Explanation :

(a) Now we have to calculate the density of mercury.

<u>Given :</u>

Volume of mercury = 25.0 ml

Mass of mercury = 340.0 g

Formula used :

\text{Density of mercury}=\frac{\text{Mass of mercury}}{\text{Volume of mercury}}

\text{Density of mercury}=\frac{340.0g}{25.0ml}=13.6g/ml

Therefore, the density of mercury is, 13.6 g/ml

(b) Now we have to calculate the mass of 120.0 ml of mercury.

As, 25.0 ml of mercury has mass = 340.0 g

So, 120.0 ml of mercury has mass = \frac{120.0ml}{25.0ml}\times 340.0g=1632g

Therefore, the mass of 120.0 ml of mercury is, 1632 grams

3 0
3 years ago
HELLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLPPPPPPPPPPPPPPPPPPPPPPP
guajiro [1.7K]
Hiii,so the problem is ‍♀️
5 0
2 years ago
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