Answer:
I'm not really sure but I thinks its
A:veuns
1) mass composition
N: 30.45%
O: 69.55%
-----------
100.00%
2) molar composition
Divide each element by its atomic mass
N: 30.45 / 14.00 = 2.175 mol
O: 69.55 / 16.00 = 4.346875
4) Find the smallest molar proportion
Divide both by the smaller number
N: 2.175 / 2.175 = 1
O: 4.346875 / 2.175 = 1.999 = 2
5) Empirical formula: NO2
6) mass of the empirical formula
14.00 + 2 * 16.00 = 46.00 g
7) Find the number of moles of the gas using the equation pV = nRT
=> n = pV / RT = (775/760) atm * 0.389 l / (0.0821 atm*l /K*mol * 273.15K)
=> n = 0.01769 moles
8) Find molar mass
molar mass = mass in grams / number of moles = 1.63 g / 0.01769 mol = 92.14 g / mol
9) Find how many times the mass of the empirical formula is contained in the molar mass
92.14 / 46.00 = 2.00
10) Multiply the subscripts of the empirical formula by the number found in the previous step
=> N2O4
Answer: N2O4
The reagents for BaCO₃ is
BaO and CO₂
<em><u>Explanation</u></em>
Reagent is a substance that bring about a chemical reaction when added to a system.
Some reagent may be added to see if a reaction has occurred.
BaO and Co₂ are reagent since they react to produce BaCO₃ as below
BaO(s) + CO₂(g) → BaCO3(s)
First , 27.6 cm^3 is equal to 27.6 ml
Density = mass g / volume cm^3
= 74.6 / 27.6 = ........ g/cm^3
Your solution in the conical flask will be acidic since you will have gone beyond the amount of acid that should completely neutralise the base completely. thax