The balanced equation that shows the reaction between oxalic acid and permanganate ion in an acidic medium is: 2MnO4- + 5H2C2O4 + 6H+ -> 2Mn(2+) + 10CO2 + 8H2O. Thus, 1 mole of oxalic acid reacts with 0.4 mole of permanganate ion. This was obtained using stoichiometry:
1 mol H2C2O4 x (2 mol MnO4-/ 5 mol H2C2O4) = 0.4 mol MnO4-
In this redox reaction, the permanganate is reduced to manganese(II) ion.
Answer:
282.7KPa
Explanation:
Step 1:
Data obtained from the question.
Number of mole of (n) = 1.5 mole
Volume (V) = 13L
Temperature (T) = 22°C = 22 + 273°C = 295K
Pressure (P) =..?
Gas constant (R) = 0.082atm.L/Kmol
Step 2:
Determination of the pressure exerted by the gas.
This can be obtained by using the ideal gas equation as follow:
PV = nRT
P = nRT /V
P = 1.5 x 0.082 x 295 / 13
P = 2.79atm.
Step 3:
Conversion of 2.79atm to KPa.
This is illustrated below:
1 atm = 101.325KPa
Therefore, 2.79atm = 2.79 x 101.325 = 282.7KPa
Therefore, the pressure exerted by the gas in KPa is 282.7KPa
Answer:
That is a compound. If it was an element it would either just be Na or Cl.
Explanation:tr
a) Molar mass of HF = 20 g/mol
Atomic mass of hydrogen = 1 g/mol
Atomic mass of fluorine = 19 g/mol
Percentage of an element in a compound:
Percentage of fluorine:
Percentage of hydrogen:
b) Mass of hydrogen in 50 grams of HF sample.
Moles of HF = 
1 mole of HF has 1 mole of hydrogen atom.
Then 2.5 moles of HF will have:
of hydrogen atom.
Mass of 2.5 moles of hydrogen atom:
1 g/mol × 2.5 mol = 2.5 g
2.5 grams of hydrogen would be present in a 50 g sample of this compound.
c) As we solved in part (a) that HF molecules has 5% of hydrogen by mass.
Then mass of hydrogen in 50 grams of HF compound we will have :
5% of 50 grams of HF = 
Answer:
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