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Answer: The coefficient in front of nitric acid is 2
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas. Liquids are represented by (l) and gases are represented by (g).
The balanced chemical reaction is :
Answer:
Bài 1. Có 30g dung dịch NaCl 20%. Tính nồng độ % dung dịch thu được khi
a. Pha thêm 20g H2O
b. Cô đặc dung dịch để chỉ còn 25g.
Answer:
The projection of the Fisher projection of D-Fructose and D-glucose is that The carbonyl carbon in D-glucose is carbon 1 (aldehyde), whereas in D-fructose, the carbonyl group is on carbon 2 (ketone).
Explanation:
An aldehyde is a compound containing a functional group with the structure −CHO, consisting of a carbonyl center and
A ketone is a functional group with the structure RC(=O)R', where R and R' can be a variety of carbon-containing substituents.
Answer:
a. 581.4 Pa
b. 3.33x10⁻⁴ mol/L
c. 3.49x10⁻⁴ mol/L
d. 0.015 g/L
Explanation:
a. By the Raoult's Law, the partial pressure of a component of a gas mixture is its composition multiplied by the total pressure, so:
pA = 0.9532*6.1
pA = 5.81452 mbar
pA = 5.814x10⁻³ bar
1 bar ----- 10000 Pa
5.814x10⁻³ bar--- pA
pA = 581.4 Pa
b. Considering the mixture as an ideal gas, let's assume the volume as 1,000 L, so by the ideal gas law, the total number of moles is:
PV = nRT
Where P is the pressure (610 Pa), V is the volume (1 m³), n is the number of moles, R is the gas constant (8.314 m³.Pa/mol.K), and T is the temperature.
n = PV/RT
n = (610*1)/(8.314*210)
n = 0.3494 mol
The number of moles of CO₂ is (V = 0.9532*1 = 0.9532 m³):
n = PV/RT
n = (581.4*0.9532)/(8.314*210)
n = 0.3174 mol
cA = n/V
cA = 0.3174/953.2
cA = 3.33x10⁻⁴ mol/L
c. c = ntotal/Vtotal
c = 0.3494/1000
c = 3.49x10⁻⁴ mol/L
d. The molar masses of the gases are:
CO₂: 44 g/mol
N₂: 28 g/mol
Ar: 40 g/mol
O₂: 32 g/mol
CO: 28 g/mol
The molar mass of the mixture is:
M = 0.9532*44 + 0.027*28 + 0.016*40 + 0.0008*28 = 43.36 g/mol
The mass concentration is the molar concentration multiplied by the molar mass:
3.49x10⁻⁴ mol/L * 43.36 g/mol
0.015 g/L
