Answer:
1st question: The moon is closer to Earth so it is easier to see and visit.
2nd question: The moon has a week/month cycle instead of a day/night cycle.
3rd question: The corona
Explanation:
Answer:
(40 g O) / (15.99943 g O/mol) x (1 mol CaCO3 / 3 mol O) x (100.0875 g CaCO3/mol) =
83 g CaCO3
So answer D), although three significant digits are not justified.
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Answer:
The volume of nitrogen oxide formed is 35.6L
Explanation:
The reaction of nitric acid with copper is:
Cu(s) + 4HNO₃ → Cu(NO₃)₂ + 2NO₂(g) + 2H₂O(l)
Moles of copper are:

Moles of nitric acid are:

As 1 mol of Cu reacts with 4 moles of HNO₃:
0.697 mol Cu × (4mol HNO₃ / 1mol Cu) = 2.79 moles of HNO₃ will react. That means Cu is limiting reactant.
Moles of NO₂ produced are:
0.697 mol Cu × (2mol NO₂ / 1mol Cu) = <em>1.394 moles of NO₂</em>
Using PV = nRT
<em>Where P is pressure (735torr / 760 = 0.967atm); n are moles (1.394mol); R is gas constant (0.082atmL/molK); T is temperature (28.2°C + 273.15 = 301.35K). </em>
Thus, volume is:
V = nRT / P
V = 1.394mol×0.082atmL/molK×301.35K / 0.967atm
V = 35.6L
<em>The volume of nitrogen oxide formed is 35.6L</em>
Explanation:
It is given that aluminium nitrate and calcium chloride are mixed together with sodium phosphate.
And,
Let us assume that the solubility be "s". And, the reaction equation is as follows.

s = 
Also, 

s = 
This means that first, aluminium phosphate will precipitate.
Now, we will calculate the concentration of phosphate when calcium phosphate starts to precipitate out using the
expression as follows.
![K_{sp} = [Ca^{2+}]^{3}[PO^{3-}_{4}]^{2}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5E%7B3%7D%5BPO%5E%7B3-%7D_%7B4%7D%5D%5E%7B2%7D)
![2.0 \times 10^{-29} = (0.016)^{3}[PO^{3-}_{4}]^{2}](https://tex.z-dn.net/?f=2.0%20%5Ctimes%2010%5E%7B-29%7D%20%3D%20%280.016%29%5E%7B3%7D%5BPO%5E%7B3-%7D_%7B4%7D%5D%5E%7B2%7D)
![2.0 \times 10^{-29} = 4.096 \times 10^{-6} \times [PO^{3-}_{4}]^{2}](https://tex.z-dn.net/?f=2.0%20%5Ctimes%2010%5E%7B-29%7D%20%3D%204.096%20%5Ctimes%2010%5E%7B-6%7D%20%5Ctimes%20%5BPO%5E%7B3-%7D_%7B4%7D%5D%5E%7B2%7D)
= 
=
M
Similarly, calculate the concentration of aluminium at this concentration of phosphate as follows.

![K_{sp} = [Al^{3+}][PO^{3-}_{4}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BAl%5E%7B3%2B%7D%5D%5BPO%5E%7B3-%7D_%7B4%7D%5D)
![9.84 \times 10^{-21} = [Al^{3+}] \times 2.21 \times 10^{-12}](https://tex.z-dn.net/?f=9.84%20%5Ctimes%2010%5E%7B-21%7D%20%3D%20%5BAl%5E%7B3%2B%7D%5D%20%5Ctimes%202.21%20%5Ctimes%2010%5E%7B-12%7D)
M
Thus, we can conclude that concentration of aluminium will be
M when calcium begins to precipitate.
1-5.037077324x10^23
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