Answer:
3.82 x 10²¹ molecules As₂O₃
Explanation:
To find the amount of molecules arsenic (III) oxide (As₂O₃), you need to (1) convert kg to lbs, then (2) convert g As₂O₃ to moles As₂O₃ (via molar mass), and then (3) convert moles to molecules (via Avogadro's number).
1 kilogram = 2.2 lb
Molar Mass (As₂O₃): 2(74.992 g/mol) + 3(15.998 g/mol)
Molar Mass (As₂O₃): 197.978 g/mol
Avogadro's Number:
6.022 x 10²³ molecules = 1 mole
0.0146 g As₂O₃ 1 kg 189 lb
------------------------ x --------------- x ------------------ x ................
1 kg 2.2 lb
1 mole 6.022 x 10²³ molecules
x ------------------ x --------------------------------------- = 3.82 x 10²¹ molecules As₂O₃
197.978 g 1 mole
Answer:
Chemical reactions involve breaking chemical bonds between reactant molecules (particles) and forming new bonds between atoms in product particles (molecules). The number of atoms before and after the chemical change is the same but the number of molecules will change.
Explanation:
Answer:
Kc = 3.90
Explanation:
CO reacts with to form and . balanced reaction is:
No. of moles of CO = 0.800 mol
No. of moles of = 2.40 mol
Volume = 8.00 L
Concentration =
Concentration of CO =
Concentration of =
Initial 0.100 0.300 0 0
equi. 0.100 -x 0.300 - 3x x x
It is given that,
at equilibrium = 0.309/8.00 = 0.0386 M
So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M
At equilibrium = 0.300 - 0.0386 × 3 = 0.184 M
At equilibrium = 0.0386 M
Answer:
Explanation:
C₂H₂ + 2H₂ = C₂H₆
1 mole 2 mole 1 mole
Feed of reactant is 1.6 mole H₂ / mole C₂H₂
or 1.6 mole of H₂ for 1 mole of C₂H₂
required ratio as per chemical reaction written above
2 mole of H₂ for 1 mole of C₂H₂
So H₂ is in short supply . Hence it is limiting reagent .
1.6 mole of H₂ will react with half of 1.6 mole or .8 mole of C₂H₂ to form .8 mole of C₂H₆
a )Calculate the stoichiometric reactant ratio = mole H₂ reacted/mole C₂H₂ reacted
= 1.6 / .8 = 2 .
b )
yield ratio = mole C₂H₆ formed / mole H₂ reacted ) = 0.8 / 1.6 = 1/2 = 0.5 .
Answer:
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Explanation:
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