This method of quantitative determination of percent purity is titrimetric reactions. These reactions most commonly involve neutralization reactions between an acid and a base. Then, we look at the neutralization reaction:
H₂C₂O₄ + 2 NaOH ⇒ Na₂C₂O₄ + 2 H₂O
So, we do the stoichiometric calculations. The important data we should know is the molar mass of oxalic acid which is equal to 90 g/mol.
(0.2283 mol/L NaOH * 0.3798 L * 1 mol H₂C₂O₄/ 2mol NaOH * 90 g/mol H₂C₂O₄) ÷ 0.7984 g *100%
= 488%
This is impossible. The purity can't be more than 100%. Looking at our calculations and the balance reaction, all steps were done correctly. So, I think there is some typographical error in the given. The mass of the sample should be 7.984 g. Then, the answer would be 48.87% purity.
Answer: The questions looks unclear
Explanation: Periodic table is a table that contains elements arranged according to their increasing atomic number.
1. D belongs to group 4
E. Belongs to group 7
B belongs to group 1
A belongs to group 8. A noble gas.
R belongs to group 3. K belongs to group 6 C belongs to group 1. H belongs to group 8
Answer : The mass of 7.0 m chain is, 15.12 kg
Explanation :
As we are given that,
The weight of the chain per unit length = 2.16 kg/m
Now we have to determine the mass of chain for 7.0 m length.
As, the mass of 1 m length of chain = 2.16 kg
So, the mass of 7.0 m length of chain = 
= 15.12 kg
Therefore, the mass of 7.0 m chain is, 15.12 kg
Answer:
PV=nRT
n = PV/RT
n = m/Mm
m/Mm = PV/RT
m = MmPV/RT
T in kelvin = T Celsius + 273.15 = 293.15 K
m = (26.04 x 1.39 x 55)/(0.08206 x 293.15)
mass in grams = 82.8 grams
Explanation:
Ideal gases formula is PV=nRT, where:
P is the pressure (1.39 atm in this case)
V is the volume (55.0 L in this case)
R is the gas constant (0.08206 L.atm/K.mole)
T is the temperature (20.0C) should be converted to Kelvin
all the unit should correspond to the one in the R.
we also know that to find the mass, we can use number mole with the formula number of mole(n) = mass (m) divided by the molar mass (Mm). therefore we substituted that in the formula and make (m) the subject of the formula.
we found the mass to be 82.8 grams
The element is Am and since you lose e- there must be a postive charge. Am+6 is the symbol
