According to avogadro constant, the number of units in one mole of any substance contain 6.022 x10 ^23 atoms
therefore the number of o atoms in one mole of CuSO4 = 6.022 x 10 ^ 23
Answer:
This would be the hydride ion where H gains an electron from a metal such as Na
Explanation:
1.
V = 200 mL (volume)
c = 3 M = 3 mol/L (concentration)
First we convert mL to L:
200 mL = 0.2 L
Then we calculate the moles using the formula: n = V × c = 0.2 L × 3 mol = 0.6 mol
Finally, we just use the molar mass of CaF2 to calculate the actual mass:
molar mass = 78 g/mol
The formula is: m = n × mm (mass = moles × molar mass)
m = 0.6 mol × 78 g/mol = 46.8 g
2.
For this question the steps are exactly like the first question.
V = 50mL = 0.05 L
c = 12 M = 12 mol/L
n = V × c = 0.05 L × 12 mol/L = 0.6 mol
molar mass (HCl) = 36.5 g/mol
m = n × mm = 0.6 mol × 36.5 g/mol = 21.9 g.
3.
The steps for this question are the opposite way.
m(K2CO3) = 250 g
molar mass = 138 g/mol
n = m ÷ mm = 1.81 mol
c = 2 mol/L
V = n ÷ c = 1.81 mol ÷ 2 mol/L = 0.905 L = 905 mL
The compound contains an ester functional group.
An ester is a carbonyl (C=O) group with an alkyl (R) group on one side and an alkoxy (OR) group on the other.
We write the <em>condensed structural formula</em> of an ester as R(C=O)OR or RCOOR.
I believe you are referring zero as the exponent. <span>Any number (except 0) with exponent 0 is defined to mean 1.
</span>
For one thing, there is a rule:
<span> a^m/ a^m = a^m-m = a^0
</span>But (when a is not equal to <span>0),
</span>
a^m/ a^m = 1
Therefore, we must define a^0 as 1.
