Answer:
The answer to your question is below
Explanation:
Trial 1 Trial 2
mass of Mg 0.255 g 0.353 g
mass of MgO 0.418 g 0.576 g
Chemical reaction
2Mg(s) + O₂(g) ⇒ 2MgO(s)
Question 1.
Atomic mass of Mg = 24.31 x 2 = 48.62 g
Molecular mass of MgO = 2(24.31 + 16) = 80.62 g
Trial 1
48.62 g of Mg ----------------- 80.62 g of MgO
0.255 g ---------------- x
x = (0.255 x 80.62)/48.62
x = 0.422 g of MgO
Trial 2 48.62 g of Mg ----------------- 80.62 g of MgO
0.353 g ---------------- x
x = (0.353 x 80.62)/48.62
x = 0.585 g of MgO
Question 2
Trial 1
Percent yield = 0.418/0.422 x 100 = 99%
Trial 2
Percent yield = 0.576/0.585 x 100 = 98.5%
Question 3
Average = (99 + 98.5)/2
= 98.75%
Hello.
balance the equation first
<span>Cu + 2AgNO3 ----> [2Ag] + Cu(NO3)2
</span>
<span>Now set up the equation </span>
<span>350gCu*(1molCu/ 63.546g Cu) * (2molAg/1molCu) * (107.8682g Ag/1molAg)
</span>
<span>Cancel out 1molCu with 1molCu, and cancel out 2molAg with 1molAg (it still remains 2mol Ag) </span>
<span>You are left with </span>
<span>350gCu*(1/63.546g Cu)*(2Ag/1)* (107.8682g Ag/1)
</span>
<span>350gCu*1*2*107.8682= 75507.74g
</span>
<span>now divide 75507.74g by 63.546g
</span>
<span>you get [1188.2374972460896g Ag]
</span>
Have a nice day
Answer:
d
Explanation:
Firstly, we need to see the theoretical mole ratio between nitrogen and ammonia from the balanced chemical equation. This is 1 to 2. One mole of nitrogen yielded two moles of ammonia.
At STP, one mole of a gas occupies a volume of 22.4, hence we need to know the volume occupied by a volume of 44.8L of ammonia. This is equal to 44.8/2 = 2 moles
Now we have seen the actual number of moles of ammonia yielded. Since this is the same as the theoretical, it means that only one mole of nitrogen was also used up.
Since it is one mole, the volume at STP is thus 22.4L
