Answer: 295.46g
Explanation: please see attached file for explanation.
<span>Infant mortality dropped significantly - raising the average age at death. It depends on how you look at the statistics. </span>
Answer:
copper(ll)oxide
Explanation:
Limit reagent is a reagent that finished thereby stopping the reaction.
2NH3(g) + 3CuO(s)---> N2(g) + 3Cu(s) + 3H20(g)
molar mass of of ammonia = 17.031 × 2 (stiochiometry mole) = 34.06 g/mol
molar mass of CuO = 79.545 × 3 = 238.635 g/mol
molar mass of Nitrogen gas = 28.0134 g/mol
34.06 g of NH3 need 238.635 of CuO to produce 28.0134 of Nitrogen gas
238.635g produce 28.0134 g
x gram of CuO produce 26 gram of Nitrogen gas
x gram = ( 26 × 238.635 ) / 28.0134 = 221.484 g
also
34.06 g of ammonia produces 28.0134g of nitogen gas
y gram of ammonia produce 26g
y gram = (26× 34.06) / 28.0134 g = 31.61 g
but the 34 g of ammonia was used in the reaction and 31.61 reacted leaving and 2.39 g of ammonia gas
The limiting reagent therefore is copper(ll)oxide
1) molar mass of the C6H8O<span>6, you need to consult the atomic weight of the C, H and O atoms that are in the periodic table: C is 12; H is 1; O is 16
(12x6)+(1x8)+(16x6)= 176g/mol
</span><span>176 g = 1 mol
0,5 g = x mol 500mg= 0,5 grams
molar mass = mass÷ moles
176 = 0,5÷ x
x= 2,84 x 10⁻³ mol
2) To find the number of molecules present in those </span>2,84 x 10⁻³ mol we need to multiple the moles by the <span>Avogadro's constant
No. of molecules = Avogadro's constant x n° of moles
</span>No. of molecules = 6.022 x 10²³ x 2,84 x 10⁻³ <span>
= 1.71 x 10²¹ molecules of vit C. </span>
I want to say B but I’m rusty on this topic
