<span>To solve this we need to balance the equations first.
So Hg + S --> HgS is balanced
One mole of Hg requires one mole of S to form one mole of HgS.
Number of moles of Sulphur = mass/ molar mass = 157/32 = 4.906
So 4.90 moles of S reacts with 4.90 moles of Hg.
Hence there are 4.90 moles of 4.90 of Hg.
Mass = number of moles * molar mass of Hg
Mass = 4.906 * 200.59 = 982.891g</span>
Answer:
1.) AgNO₃
2.) 0.563 moles AgBr
Explanation:
The limiting reagent is the reagent that is used up completely during a reaction. It can be identified by calculating which reactant produces the smallest amount of product. This can be done by determining the number of moles of each reagent (via molarity conversion). and then converting it to moles of the product (via mole-to-mole ratio).
AgNO₃ (aq) + KBr (aq) ---> AgBr (s) + KNO₃ (aq)
Molarity (M) = moles / liters
100 mL = 1 L
AgNO₃
45.0 mL / 100 = 45.0 L
1.25 M = ? moles / 0.450 L
? moles = 0.563 moles
KBr
75.0 mL / 100 = 0.750 L
0.800 M = ? moles / 0.750 L
? moles = 0.600 moles
In this case, there is no need to use the mole-to-mole ratio because all of the coefficients are one in the reaction (the amount of the limiting reagent used is the same amount of product produced). Since AgNO₃ produces the smaller amount of product, it is the limiting reagent.
Answer:
C. 0.4.
Explanation:
<em>∵ mole fraction of acetic acid (X acetic acid) = (no. of moles acetic acid)/(total no. of moles) = (no. of moles acetic acid)/(no. of moles of acetic acid + no. of moles of water).</em>
<em></em>
- no. of moles of acetic acid = 2, no. of moles of water = 3.
- Total no. of moles = no. of moles of acetic acid + no. of moles of water = 2 + 3 = 5.
<em>∴ mole fraction of acetic acid (X acetic acid) = (no. of moles acetic acid)/(total no. of moles) =</em> (2)/(5)<em> = 0.4.</em>
In one mole of glucose 38 ATP energy is stored this accounts for only 40 per-cent of the total energy in glucose.
Explanation:
In standard conditions, during the cellular respiration 1 mole of Glucose in the presence of oxygen produces 36 or 38 ATPs. This accounts for only 40% of the total energy as the remaining 60 per-cent of the energy is dissipated as heat.
I mole of glucose enters the glycolysis step of aerobic cellular respiration which after oxidative phosphorylation and Electron transport chain would give 38 ATP molecules.
It can be said that only 38.3% of energy is put in ATP molecules.
I think it could be C maybeee though
