this is a dilution equation where 50.0 mL of 1.50 M H₂SO₄ is taken and added to 200 mL of water.
c1v1 = c2v2
where c1 is concentration and v1 is volume of the concentrated solution
and c2 is concentration and v2 is volume of the diluted solution to be prepared
50.0 mL of 1.50 M H₂SO₄ is added to 200 mL of water so the final solution volume is - 200 + 50.0 = 250 mL
substituting these values in the formula
1.50 M x 50.0 mL = C x 250 mL
C = 0.300 M
concentration of the final solution is A) 0.300 M
Two precursor alkenes
H₃C CH₃
I I
H₂C=C-CH-CH₃ 2,3-dimethyl-1-butene
H₃C CH₃
I I
H₃C-CH=CH-CH₃ 2,3-dimethyl-2-butene
alkane
H₃C CH₃
I I
H₃C-CH-CH-CH₃ 2,3-dimethylbutane
H₃C CH₃ H₃C CH₃
I I I I
H₂C=C-CH-CH₃ + H₂ → H₃C-CH-CH-CH₃
H₃C CH₃ H₃C CH₃
I I I I
H₂C-C=CH-CH₃ + H₂ → H₃C-CH-CH-CH₃
Answer:
The sum of the protons and neutrons.
Explanation:
protons + neutrons = atomic mass
Answer:
Equal volumes of SO2(g) and O2(g) at STP contain the same number of molecules
Explanation:
According to Avogadro Law,
Equal volume of all the gases at same temperature and pressure have equal number of molecules.
This law state that volume and number of moles of gas have direct relation.
When the amount of gas increases its volume will increase and when the amount of gas decreases its volume will decrease.
Mathematical relation:
V ∝ n
V/n = K
K is proportionality constant.
When number of moles change from n₁ to n₂ and volume from V₁ to V₂
expression will be,
V₁/n₁ = K , V₂/n₂ = K
V₁/n₁ = V₂/n₂
Answer:
In 1 mol of Pb₃(PO₄)₄ occupies 1001.48 grams
Explanation:
This compound is the lead (IV) phosphate.
Grams that occupy 1 mole, means the molar mass of the compound
Pb = 207.2 .3 = 621.6 g/m
P = 30.97 .4 = 123.88 g/m
O = (16 . 4) . 4 = 256 g/m
621.6 g/m + 123.88 g/m + 256 g/m = 1001.48 g/m
