Answer:
5 mL
Explanation:
Given data:
mass of ring = 107 g
volume of water = 10 mL
increase in volume = 15 mL
How much water displace = ?
Solution:
V (ring) = V (water + ring) - V (water)
V (ring) = 15 mL - 10 mL
V (ring) = 5 mL
when the ring is put into cylinder, volume is increased by 15 mL. The volume of water was 10 mL so water is displaced by 5 mL and the volume 5mL is the voulme of ring.
B. a circle graph
circle graphs are the best to show percentages because they’re very easy to look at and get info from
The given chemical reaction given above is already balanced such that the number of atoms in the left hand side of the equation is equal to that of the right hand side. Using the dimensional analysis, proper conversion factors and the molar masses,
mass of nitrogen = (0.129 g H₂)(1 mol H₂/2 g H₂)(1 mol N₂/3 mol H₂)(28 g N₂/1 mol N₂)
mass of nitrogen = 0.602 g N₂
Therefore, 0.602 g of nitrogen will be required for he reaction.
Molar mass of CaCl2 = 40+ ( 35.5 ×2)=110
Mr of Ca(OH)2 = 40+ (16+1)×2 =74
%of Ca = (40÷ 74)×10=...
1 m = 100cm...
1cm = (1÷100) m
So 45.5 cm = 45.5 ×(1÷100) =....
1km = 1000m
1m = 100 cm
1cm =10mm
So 1km = 1000×100×10 mm
Now convert
The volume of CO2 at STP =124.298 L
<h3>Further explanation</h3>
Given
Reaction
4 KMnO4, +4 C3H5(OH)5, -7K2CO3, + 7 Mn2O3, +5 CO2, + 16 H2O
701,52 g of KMnO4
Required
volume of CO2 at STP
Solution
mol KMnO4 (MW=158,034 g/mol) :
mol = mass : MW
mol = 701.52 : 158.034
mol = 4.439
mol CO2 from equation : 5/4 x mol KMnO4 = 5/4 x 4.439 = 5.549
At STP 1 mol = 22.4 L, so for 5.549 moles :
=5.549 x 22.4
=124.298 L
