Remember the acronym "Oil Rig". Oxidation is loss, Reduction is gain of electrons. Calcium is losing electrons so it's an oxidation reaction.
Answer:
S = 21.92 %
F = 78.08 %
Explanation:
To find the percent composition of each element in SF6, we must find the molar mass of SF6 first.
Molar mass of SF6 = 32 + 19(6)
= 32 + 114
= 146g/mol
mass of Sulphur (S) in SF6 = 32g
mass of Fluorine (F) in SF6 = 114g
Percent composition = mass of element/molar mass of compound × 100
- % composition of S = 32/146 × 100 = 21.92%.
- % composition of F = 114/146 × 100 = 78.08%.
Answer:
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Explanation:
<u>Step 1: </u>Data given
mass of water = 300 grams
initial temperature = 10°C
final temperature = 50°C
Temperature rise = 50 °C - 10 °C = 40 °C
Specific heat capacity of water = 4.184 J/g °C
<u>Step 2:</u> Calculate the heat
Q = m*c*ΔT
Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)
Q = 50208 Joule = 50.2 kJ
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
One thing to notice in the question is, we are asked about molecular oxygen that has formula O2 not atomic oxygen O.
As we are asked about molecular oxygen, we will answer the question in terms of number of molecules that are present in 16 grams of molecular oxygen.
To get the number of molecules present in 16 grams of O2, we will use the formula:
No. of molecules = no. of moles x Avogadro's number (NA)----- eq 1)
As we know:
The number of moles = mass/ molar mass of molecule
Here we have been given mass already, 16 grams and the molar mass of O2 is 32 grams.
Putting the values in above formula:
= 16/32
= 0.5 moles
Putting the number of moles and Avogadro's number (6.02 * 10^23) in eq 1
No. of molecules = 0.5 x 6.02 * 10^23
=3.01 x 10^23 molecules
or 301,000,000,000,000,000,000,000 molecules
This means that 16 grams of 3.01 x 10^23 molecules of oxygen.
Hope it helps!
Answer: 0.055 moles of 
are produced by the reaction of 0.055 mol of ammonium perchlorate.
Explanation:
The balanced chemical reaction for decomposition of ammonium perchlorate is:
According to stoichiometry :
2 moles of 
produce = 2 moles of
Thus 0.055 moles of will produce =
of
Thus 0.055 moles of are produced by the reaction of 0.055mol of ammonium perchlorate.
