Answer:
1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride? atoms of boron.
2. How many MOLES of fluorine are present in of boron trifluoride? moles of fluorine.
Explanation:
The molecular formula of boron trifluoride is 
.
So, one mole of boron trifluoride has one mole of boron atoms.
1. The number of boron atoms in 2.20 moles of boron trifluoride is 2.20 moles.
The number of atoms in 2.20 moles of boron is:
One mole of boron has ---- 
atoms.
Then, 2.20 moles of boron has
-
2. Calculate the number of moles of BF3 in 5.35*1022 molecules.
One mole of boron trifluoride has three moles of fluorine atoms.
Hence, 0.0888moles of BF3 has 3x0.0888mol of fluorine atoms.
=0.266mol of fluorine atoms.
PH (potential of hydrogen) is a numeric scale that is used to show the acidity or basicity of an aqueous solution. It tells how acidic or alkaline a substance is . The pH values ranges from 0 to 14, such that acidic solutions have values between 1 to 6.9 with most acidic having a pH value of 1 and those that are basic have values from 7.1 to 14, with most acidic having a value of 14. Acidic compounds contain replaceable hydrogen ions while basic compounds contain hrdroxyl ions. In this case, a coke has a pH of 3.5 (acidic) which means that it has an excess of hydrogen ions (H+) and would be called an acid.
It is.
An acid will be strong when its conjugated base is highly stable, and vice-versa.
That can occur for instance through electronic delocalization.
1) is sulfuric acid
2)is nitrous acid
3)is hydrochlorous acid
4)is hydrobromous acid
5)is hydrophosphoric acid
6)is fluoric acid
7)is sulfuric acid
8)is chlorous acid
9)is nitric acid
10)is iodic acid
11)is acetous acid
12)is chlorous acid.
Answer:
48.8%
Explanation:
The reaction has a 1:1 mole ratio so;
Number of moles of benzoic acid reacted = mass/molar mass = 3.8 g/122.12 g/mol = 0.03 moles
So;
0.03 moles of methyl benzoate is formed in the reaction
Mass of methyl benzoate formed = 0.03 moles * 136.15 g/mol = 4.1 g
percent yield = actual yield/theoretical yield * 100/1
percent yield = 2.0 g/4.1 g * 100 = 48.8%
