The formula unit for Li and O is —————-> Li2O
Answer:
C. CH3COOH, Ka = 1.8 E-5
Explanation:
analyzing the pKa of the given acids:
∴ pKa = - Log Ka
A. pKa = - Log (1.0 E-3 ) = 3
B. pKa = - Log (2.9 E-4) = 3.54
C. pKa = - Log (1.8 E-5) = 4.745
D. pKa = - Log (4.0 E-6) = 5.397
E. pKa = - Log (2.3 E-9) = 8.638
We choose the (C) acid since its pKa close to the expected pH.
⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):
- pH = pKa + Log ([CH3COO-]/[CH3COOH])
∴ pH = 4.5
∴ pKa = 4.745
⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])
⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])
⇒ 0.5692 = [CH3COO-]/[CH3COOH]
∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-])/[CH3COOH]
⇒ 1.8 E-5 = [H3O+](0.5692)
⇒ [H3O+] = 3.1623 E-5 M
⇒ pH = - Log ( 3.1623 E-5 ) = 4.5
Answer: The molarity of solution is 0.231 M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
where,
n = moles of solute
= volume of solution in L
Molar mass of 
= 
moles of = 
Now put all the given values in the formula of molality, we get
Therefore, the molarity of solution is 0.231 M
Answer:
6.2moles of Gold
Explanation:
To solve this problem, we are going to use the mole concept approach.
Given that;
Number of atoms of gold is 3.73 x 10²⁴ atoms
Now;
In 1 mole of any substance, we have 6.02 x 10²³ atoms;
So;
If there 6.02 x 10²³ atoms in 1 mole of any substance;
3.73 x 10²⁴ atoms will contain 
= 6.2moles of Gold
Here's three i can name off the top of my head.
Physics , Chemistry and Biology are all real types of science.
