How does the law of conservation of mass apply to this reaction: C2H4 + O2 → H2O + CO2?
Answer:
a) T
b) T
c) F
d) F
e) T
f) T
g) T
h) F
I) F
j) F
k) F
l) F
Explanation:
The w/v concentration is obtained from, mass/volume. Hence;
%w/v= 50/1000= 5%
In the %w/w we have;
25g/100 g = 25% w/w
In combustion reaction, energy is given out hence it is exothermic.
Neutralization reaction yields a salt and water
% by mass of carbon is obtained from;
8× 12/114 × 100 = 84.1%
All the ionic substances mentioned have very low solubility in water.
One mole of a substance contains the Avogadro's number of each atom in the compound.
There are two iron atoms so one mole contains 2× 55.85 g of iron.
Some sulphates such as BaSO4 are insoluble in water.
Halides are soluble in water hence NaI is soluble in water.
The equation does not balance with the given coefficients because the number of atoms of each element on both sides differ.
The equation represents a decomposition of calcium carbonate as written.
Answer:
6.9 (two sig figs)
Explanation:
2.375 + 4.5 = 6.875 = 6.9
When adding or subtracting, sig figs are determined by the least number of digits past the decimal point.
Answer:
Let me give it a try.
H3PO4 + Ca(OH)2 = Ca3(PO4)2 + H2O
Balancing this reaction
2H3PO4 + 3Ca(OH)2 == Ca3(PO4)2 + 6H2O.
Moles= Molarity x Volume
Volume = 38.5ml = 0.0385L
Moles of Ca hydroxide = 0.150m/L x 0.0385L
(Notice the units canceling out...leaving moles).
=0.005775moles of Ca(OH)2.
From balanced reaction...
3moles of Ca(OH)2 completely reacts with 2moles of H3PO4
0.005775moles of Ca(OH)2 would completely react with....
= 0.005775 x 2/(3)
=0.00385moles of H3PO4.
Now we're looking for its Concentration in Mol/L
Molarity=Moles of solute/Volume of solution(in L)
Volume of solution assuming no other additions to the reaction = 15ml + 38.5ml =53.5ml =0.0535L
Molarity = 0.00385/0.0535
=0.072Mol/L.
If this is wrong
then Simply Try The formula for Mixing of solutions
C1V1 = C2V2
0.15 x 38.5 = C2 x (15+38.5)
C2 = 0.11M/L.
Answer:
Carbohydrates,Monosaccharides,and disaccharides
Explanation: I'm good at science
