20.06 g of Hg and 1.6 g of O₂
<u>Explanation:</u>
To Find:
Number of Mercury and oxygen that can be obtained from 21.7 g of HgO
First we have to write the balanced equation for the decomposition reaction of Mercury(II) oxide as,
2 HgO (s) → 2Hg(l) + O₂ (g)
21.7 g of HgO = 
= 0.1 mol of HgO.
As per the above equation, we can find the mole ratio between HgO and Hg is 1: 1 and that of HgO and oxygen is 2:1 .
So amount of Hg produced = 0.1 mol × 200.59 g / mol ( molar mass of Hg)
= 20.06 g of Hg
Amount of oxygen produced = 0.05 mol × 32 g/ mol = 1.6 g of O₂
Thus it is clear that 20.06 g of Hg and 1.6 g of O₂ is obtained from 21.7 g of HgO
D.
This is in accordance with the law of conservation of energy.
Answer:
hope its not to late..............Samira's model correctly demonstrates how the properties changed with the rearrangement of the atoms. However not all atoms are accounted for. There is a missing reactant. Samira's model correctly demonstrated how the atoms in two compounds reacted to form two new products. However, the elements present in the reactants side should be the elements that make up the new products in the product side. But as the diagram shows, Sameera has mistakenly added a new element to one of her products which will be wrong.
Explanation:
Answer:
2Al+ 6HNO3 ---- 3H2 + 2Al(NO3)3
Explanation:
Put coefficient a,b,c, and d for calculation:
a Al + b HNO3 = c H2 + d Al(NO3)3
for Al: a = d
for H: b = 2c
for N: b = 3d
for O: 3b = 9d
Suppose a=1, then d=1, b=3, c=3/2
multiply 2 to make all natural number, a=2, then b=6, c=3, d=2
A catalyst is a chemical substance that hastens the chemical reaction. This does not participates in the creating the product(s) but allows it to be formed easily. With this, it is now known that the rate of the reaction becomes relatively higher compared to the uncatalyzed reactions.
Therefore, the answer to this item is the rate of the reaction becomes faster.
