Answer:
100ml of a stock 50% KNO3 solutions are needed to prepare 250ml of a 20% KNO3 solution.
Explanation:
In the given question it is mentioned that
S1=50%
V2=250ml
S2= 20%
We all know that
V1S1=V2S2
∴V1= V2×S2÷S1
∴V1= V2S2×1/S1
∴V1= 250×20÷50
∴V1= 100ml
Answer:
-125.4
Explanation:
Target equation is 4C(s) + 5H2(g) = C4H10
These are the data equations for enthalpy of combustion
- C(s) + O2(g) =O2(g) -393.5 kJ/mol * 4
- H2(g) + ½O2(g) =H20(l) = 285.8 kJ/mol * 5
- 2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 - 2877.1 reverse
To get target equation multiply data equation 1 by 4; multiply equation 2 by 5; and reverse equation 3, so...
Calculate 4(-393.5) + 5(-285.8) + 2877.6 and you should get the answer.
Answer:
2Li + F₂ → 2LiF
Explanation:
The reaction expression is given as:
Li + F₂ → LiF
We are to balance the expression. In that case, the number of atoms on both sides of the expression must be the same.
Let use a mathematical approach to solve this problem;
Assign variables a,b and c as the coefficients that will balance the expression:
aLi + bF₂ → cLiF
Conserving Li: a = c
F: 2b = c
let a = 1, c = 1 and b = 
Multiply through by 2;
a = 2, b = 1 and c = 2
2Li + F₂ → 2LiF
Zn, Cd, and Ag are transition metals that usually form only one monoatomic cation.
A monatomic cation is a cation made of only one atom.
Cations are positively charged ions, in this example Ag⁺, Cd²⁺ and Zn²⁺.
These cations form only one type of ion, while iron and copper form more than one type of cations.
Iron and copper form cations with different charges (Fe²⁺, Fe³⁺, Cu⁺, Cu²⁺).
It depends on electron configuration which type would be formed.
Electron configuration of zinc atom: ₃₀Zn 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²
Transition metals are elements in the d-block of the Periodic table.
More about transition metals: brainly.com/question/12843347
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Colligative properties include lowering the freezing point, lowering the vapor pressure, and raising the boiling point.
