PLEASE HELP!!!!! Calcium hydroxide is _____. an acid a base a neutral
2 answers:
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Answer:
The answer to your question is V₁ = 12.5 ml
Explanation:
Data
Volume = V₁?
[NaOH] = C₁ = 4.0 M
Volume 2 = V₂ = 100 ml
[NaOH] = C₂ = 0.5 M
Formula of dilution
V₁C₁ = V₂C₂
Solve for V₁ (original solution)
V₁ =
Substitution
V₁ =
Simplification
V₁ =
Result
V₁ = 12.5 ml
Data:
M (molarity) = ? (M or Mol/L)
m (mass) = 13.50 g
V (volume) = 250 mL → 0.25 L
MM (Molar Mass) of Lead(IV) Nitrate
Pb = 1*207 = 207 amu
N = (1*14)*4 = 14*4 = 56 amu
O = (3*16)*4 = 48*4 = 192 amu
------------------------------------
MM of = 207+56+192 = 455 g/mol
Formula:
Solving:
Answer:
<span>B. 0.119 M</span>
M (molarity) = ? (M or Mol/L)
m (mass) = 13.50 g
V (volume) = 250 mL → 0.25 L
MM (Molar Mass) of Lead(IV) Nitrate
Pb = 1*207 = 207 amu
N = (1*14)*4 = 14*4 = 56 amu
O = (3*16)*4 = 48*4 = 192 amu
------------------------------------
MM of
Formula:
Solving:
Answer:
<span>B. 0.119 M</span>
Answer:
a) LAW OF MULTIPLE PROPORTIONS
b) 0.095g, 0.71g, 0.285g respectively
c) X2Y, YZ15, X6Y
d) hence mass of compound X = 21 x 0.045 = 0.95g
mass of compound Y = 21 x 0.955 = 20.05g
Explanation:
a) The assumptions made in solving this questions is the application of the LAW OF MULTIPLE PROPORTIONS. The Law of multiple proportions states that if two elements A and B combine together to form more than one compound, then the several masses of A which chemically combine with a fixed mass of B is in a simple ratio.
for example, copper forms two oxides ; copper(I) oxide (CuO) and copper(ii) oxide(Cu2O), it is possible for the two samples of the oxides to be reduced to Cu by reacting with Hydrogen gas. as such, certain masses of oxygen combine separately with a fixed mass of Cu. then the ratios of Cu are then determined.
b) To calculate the relative masses, we take note of the three compounds given, they all have some amount of Y in them, hence we can use Y as our relative mass, this implies that the relative mass of Y = 1g
mass of X = 0.4g
mass of Y = 4.2g
amount of X in 1g of Y = 0.4 x 1 /4.2
= 0.095g
for compound 2;
mass of Y = 1.4g
mass of Z = 1.0g
amount of Z in 1g of Y =1.0 x 1 /1.4
= 0.71g
for compound 3;
mass of X = 2.0g
mass of Y = 7.0g
amount of X in 1g of Y = 1 x 2/7
= 0.285g
c) Applying the law of multiple proportions; since elements X and Z combine with a fixed mass of Y, they must bear a simple ratio;
compound 1/compound 3 = 0.095/0.285
= 1/3
compound 1/compound 2 = 0.095/0.71
= 2/15
compound 2/ compound 3 = 0.71/0.285
= 5/2
formular for compound 1 = X2Y
formula for compound 2 = YZ15
formular for compound 3 = X6Y
d) from the formular X2Y, we can get the amount of each product in XY using the ratios
%of compound XY in X = mass of compound X / total Mass
= 0.2/4.4 = 4.5%
as such in a 21g of compound XY, %of compound Y = 1 - %of compound X = 95.5%
hence mass of compound X = 21 x 0.045 = 0.95g
mass of compound Y = 21 x 0.955 = 20.05g
