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Answer:
0.676 grams of manganese (IV) oxide should be added.
Explanation:
Moles of chlorine gas = n
Volume of the chlorine gas = V = 205 mL = 0.205 L
Pressure of the chlorine gas = 705 Torr =
1 atm = 760 Torr
Temperature of the chlorine gas = T = 25°C = 25 + 273 K = 298 K
( ideal gs equation)
According to reaction, 1 mole of chlorine gas is obtained from 1 mole of manganese(IV) oxide,then 0.00777 moles of chlorine gas will be obtained from :
of manganese (IV) oxide
Mass of 0.00777 moles of manganese (IV) oxide:
0.00777 mol × 87 g/mol = 0.676 g
0.676 grams of manganese (IV) oxide should be added.
Answer: (a) There are 0.428 moles present in 12 g of molecule.
(b) There are 2 moles present in particles of oxygen.
Explanation:
(a). The mass of nitrogen molecule is given as 12 g.
As the molar mass of is 28 g/mol so its number of moles are calculated as follows.
So, there are 0.428 moles present in 12 g of molecule.
(b). According to the mole concept, 1 mole of every substance contains atoms.
Therefore, moles present in particles are calculated as follows.
So, there are 2 moles present in particles of oxygen.