Answer:
Pure water has an acidity of about 7 on the pH scale. -is a chemical property of pure water. Pure water has an acidity of about 7 on the pH scale
So,
Formate has a resonating double bond.
In molecular orbital theory, the resonating electrons are actually delocalized and are shared between the two oxygens. So the carbon-oxygen bonds can be described as 1.5-bonds (option B). I'm not sure if option C is correct, however, because the likelihood of both delocalized electrons being in the area of one oxygen atom is less than 50%.<span />
The atomic mass would be 28.08535 amu. Multiply 27.9769 by .92297 = 25.803. Multiply 28.9765 by .046832 to get 1.357. Multiply 29.9738 by .03872 to get .925351136. Add 25.803 + 1.357 + .03872 to get 28.08535 amu
Answer:
Explanation:
To convert from moles to grams, the molar mass must be used.
1. Find Molar Mass
The compound is iron (III) chloride: FeCl₃
First, find the molar masses of the individual elements in the compound: iron (Fe) and chlorine (Cl).
There are 3 atoms of chlorine, denoted by the subscript after Cl. Multiply the molar mass of chlorine by 3 and add iron's molar mass.
- FeCl₃: 3(35.45 g/mol)+(55.84 g/mol)=162.19 g/mol
This number tells us the grams of FeCl₃ in 1 mole.
2. Calculate Moles
Use the number as a ratio.
Multiply by the given number of grams, 345.0.
Flip the fraction so the grams of FeCl₃ will cancel.
Divide.
3. Round
The original measurement of grams, 345.0, has 4 significant figures. We must round our answer to 4 sig figs.
For the answer we calculated, that is the thousandth place.
The 1 in the ten thousandth place tells us to leave the 7 in the thousandth place.
There are about <u>2.127 mole</u>s of iron (III) chloride in 345.0 grams.
Answer:
92.49 %
Explanation:
We first calculate the number of moles n of AgBr in 0.7127 g
n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g
n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol
Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and
From n = m/M
m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol
m = 0.0038 mol × 79.904 g/mol = 0.3036 g
% Br in compound = m₁/m₂ × 100%
m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)
m₂ = mass of compound = 0.3283 g
% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %
