Answer:
It depends on where it's located on the periodic table.
Explanation:
Metals are on left side of the periodic table except Hydrogen (H), which is a nonmetal, then there's the staircase, which has the metalloids, and on the right of the staircase are the nonmetals.
Answer:
Second Earth's Atmosphere
The second atmosphere, which was the first to remain with the earth, was created by volcanic outgassing and comet ice. There was a lot of water vapor, carbon dioxide, phosphorus, and methane in this atmosphere, but absolutely no oxygen.
Explanation:
Answer:
56.69905
Explanation:
The conversion factor from pounds to kilograms is 0.45359237, which means that 1 pound is equal to 0.45359237 kilograms.
1. you need a periodic table and find the atomic mass of Cu (copper), S (sulfur) and O (oxygen). The atomic mass is the number in the box that corresponds to the element and have several decimal places.
2. atomic mass of
Cu = 63.546
S = 32.065
O = 15.9994
3. Then according to the formula of the compound, you add as many time the atomic mass of each element as subindex in the formula and add all the values together to calculate the molecular mass of the compound in grams.
4. 63.546 g + 32.065 g + ( 4 x <span>15.9994) = 159.609 g
5. this value </span><span>159.609 g is the mass in grams of one mol of CuSO4
6 the problem is asking not for the mass of one mole but the mass of 3.65 moles of CuSO4
7 then you have the multiply the value of one mol by the number of moles that the problem is asking you
8. </span><span>159.609 g x 3.65 = 582.571 g
</span>
9 the answer to the problem will be
"there are 582.571 g of CuSO4 in 3.65 moles of CuSO4"
pH of solution = 13.033
<h3>Further explanation</h3>
Given
2.31 g Ba(OH)₂
250 ml water
Required
pH of solution
Solution
Barium hydroxide is fully ionized, means that Ba(OH)₂ is a strong base
So we use a strong base formula to find the pH
[OH ⁻] = b. Mb where
b = number of OH⁻
/base valence
Mb = strong base concentration
Molarity of Ba(OH)₂(MW=171.34 g/mol) :
Ba(OH)₂ ⇒ Ba²⁺ + 2OH⁻(b=valence=2)
[OH⁻]= 2 . 0.054
[OH⁻] = 0.108
pOH= - log 0.108
pOH=0.967
pOH+pH=14
pH=14-0.967
pH=13.033
