Given :
A 10.99 g sample of NaBr contains 22.34% Na by mass.
To Find :
How many grams of sodium does a 9.77g sample of sodium bromine contain.
Solution :
By law of constant composition , in any given chemical compound, the elements always combine in the same proportion with each other.
Therefore , percentage of Na by mass in NaBr will be same for every amount .
Percentage of Na in 9.77 g NaBr is 22.34 % too .
Gram of Na = 
.
Hence , this is the required solution .
Answer:
'An ion has a non-zero electric charge. A radical has an atom with unfilled electron shells and so is very reactive, but is electrically neutral.'
'Atoms are single neutral particles. Molecules are neutral particles made of two or more atoms bonded together.'
'The primary difference that lies between these organic compounds and inorganic compounds is that organic compounds always have a carbon atom while most of the inorganic compounds do not contain the carbon atom in them.'
B. the frogs are a limiting factor for the gnats
the frogs limit the reproduction of the gnats, and therefore with less frogs the gnat population can increase
88.98 %
The Balance Chemical Equation is as follow,
2 HCl + Pb(NO₃)₂ → 2 HNO₃ + PbCl₂
According to equation,
331.2 g (1 mole) Pb(NO₃)₂ produces = 278.1 g (1 mole) PbCl₂
So,
870 g of Pb(NO₃)₂ will produce = X g of PbCl₂
Solving for X,
X = (870 g × 278.1 g) ÷ 331.2 g
X = 730.5 g of PbCl₂
Therefore,
Theoretical Yield = 730.5 g
Also as given,
Actual Yield = 650 g
So using following formula for percentage yield,
%age Yield = (Actual Yield / Theoretical Yield) × 100
Putting values,
%age Yield = (650 g / 730.5 g) × 100
%age Yield = 88.98 %
Brianliest please and thank you.
Answer:
1.16L can be made
Explanation:
Molarity = Mol / Volume
Volume = Mol / Molarity
Let's determine the moles of salt, with that mass:
130 g FeCl₂ . 1mol / 126.75 g = 1.02 moles of FeCl₂
Volume = 1.02 mol / 0.88 mol/L → 1.16L
