Answer:
water has a high specific heat
Explanation:
this should work as an answer
Answer: grams=0.048g, ounces=0.0017oz, 0.00011lb
Explanation:
Stoichiometry
48 mg x 1 g
÷ 1000 mg = 0.048 g
48 mg x 1 g x 16 oz
÷ 1000 mg ÷ 453.6 g = 0.0017 oz
48 mg x 1 g x 1 lb
÷ 1000 mg ÷ 453.6 g = 0.00011 lb
Answer: 1.24 × 10^25
Explanation:
× 
Using our knowledge in unit conversions, we know the mole units cancel each other out and all there's left is the atom unit. From here we can multiply the fractions and eventually we end with the number 124.0532 × 10^23
According to the scientific notation rules, the number to the left of the decimal cannot exceed 10 so we have to move the decimal to the left two spaces. With this change, we also have to change the exponent of the 10. Because we moved the decimal point two spaces to the left, that means we have 10^25.
Explanation:
In this reaction, the reactants are Li and N2. The product is Li3N
So we have;
Li + N2 → Li3N
Upon balancing, we have;
6Li + N2 → 2 Li3N
The sum of the coefficients is 6 + 1 + 2 = 9
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:
