Answer:535.5g
Explanation:
1) find the volume of the object
=L*W*H
=1.8*3.5*10.0
=63cm^3
2) the original formula is solving for the dencity which we have so...
D=M/V
M=D*V
M=8.5(63)
M=535.5g
<u>Ans: O2 because it produces only 0.20 mol of NO</u>
<u>Given:</u>
Mass of NH3 = 4.0 g
Mass of O2 = 8.0 g
<u>To determine:</u>
The limiting reagent in the given reaction
4NH3 + 5O2 → 4NO + 6H2O
<u>Explanation:</u>
# moles of NH3 = 4.0/17 g.mol-1 = 0.2353 moles
# moles of O2 = 8.0/32 = 0.25 moles
Based on the reaction stoichiometry:
1 mole of NH3 produces 1 mole of NO
Therefore, 0.2353 moles of NH3 will produce: 0.2353 moles of NO
Similarly, 1 mole of O2 will produce 4/5 moles of NO
0.25 moles of O2 will form: 0.25*4/5 = 0.2 moles of NO
Thus, O2 is the limiting reactant.
Answer: The molecular formula will be 
Explanation:
Mass of C= 49.47 g
Mass of N = 28.85 g
Mass of O = 16.48 g
Mass of H = 5.20 g
Step 1 : convert given masses into moles.
Moles of C =
Moles of N =
Moles of O =
Moles of H =
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = 
For N =
For O =
For H =
The ratio of C : N: O: H = 4: 2: 1: 5
Hence the empirical formula is
The empirical weight of
= 4(12)+2(14)+1(16)+5(1)= 97 g.
The molecular weight = 194.19 g/mole
Now we have to calculate the molecular formula.
The molecular formula will be = 
Answer:
C + O2 → CO2
Explanation:
C + O2 → CO ----------------- (1)
from equ (1) on reactant side, C has 1 mole, O has 2 moles
from equ (1) on product side, C has 1 mole, O has 1 mole
Thus, to balance the equation, O should have 2 moles
C + O2 → CO2
Answer:
C. A solution that cannot dissolve any more solute.
Explanation:
hope it helps .