Answer:
D) B and D. Weak tides, called neap tides, are experienced in the Earth's oceans when the sun is in positions B or D. In this case, the sun and moon interfere with each other in producing tidal bulges on Earth. It is definitely not the other answer since I got it right on USATESTPREP.
Answer:
18.76atm
Explanation:
Using the formula V1P1/T1 = V2P2/T2, from combined gas law. Volume is constant since we have not been given. Therefore the formula comes to be; P1/T1 = P2/T1
To get P2 = T2(P1/T1)
Where P2 is final pressure
P2 = 239K ( 23atm/293K)
=18.76atm
Answer:
A. 2
hope this helps :)
if you need an explanation let me know!!
2 moles of sodium hydroxide will be needed.
<h3><u>Explanation</u>:</h3>
Sodium hydroxide is a compound which is a base and nitric acid is the acid. The formula of the nitric acid is HNO3 and that of sodium hydroxide is NaOH.
The reaction between them are
NaOH +HNO3 =NaNO3 +H2O.
So here we can see that 1 mole of sodium hydroxide reacts with 1 mole of nitric acid to produce 1 mole of sodium nitrate and 1 mole of water.
So for 2 moles of nitric acid, 2 moles of sodium hydroxide will be required.
Answer:
4.96E-8 moles of Cu(OH)2
Explanation:
Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.
Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.
![pH= -log[H]\\pH= -log (\frac{kw}{[OH]})](https://tex.z-dn.net/?f=pH%3D%20-log%5BH%5D%5C%5CpH%3D%20-log%20%28%5Cfrac%7Bkw%7D%7B%5BOH%5D%7D%29)
![8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}](https://tex.z-dn.net/?f=8.23%20%3D%20-%20log%28%5Cfrac%7BKw%7D%7B%5BOH%5D%7D%20%5C%5C10%5E%7B-8.23%7D%20%3D%20Kw%2F%5BOH%5D%5C%5C%20%5BOH%5D%20%3D%20Kw%2F10%5E%7B-8.23%7D)
![[OH]=1.69E-6](https://tex.z-dn.net/?f=%5BOH%5D%3D1.69E-6)
This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":
The expression for Kps is:
![Kps= [Cu^{2+}] [OH]^2](https://tex.z-dn.net/?f=Kps%3D%20%5BCu%5E%7B2%2B%7D%5D%20%5BOH%5D%5E2)
The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

"s" is the soluble quantity of Cu(OH)2.
The solution for this third grade equation is 
Now, let us calculate the moles in 1 L:
