Answer:
I don't understand what you are asking
Answer:
[NH₃] = 14.7 mol/L
Explanation:
28 wt% is a type of concentration that indicates that 28 g of ammonia is contained in 100 g of solution.
Let's determine the amount of ammonia:
28 g . 1 mol / 17.03g = 1.64 moles of NH₃
You need to consider that, when you have density's data it is always referred to solution:
Mass of solution is 100 g, let's find out the volume
0.90 g/mL = 100 g /V
V = 100 g / 0.90mL/g → 111.1 mL
We convert the volume to L → 111.1 mL . 1 L/1000mL = 0.1111 L
mol/L = 1.64 mol/0.1111L → 14.7 M
mol/L = M → molarity a sort of concentration that indicates the moles of solute in 1L of solution
Answer:
Kr is a Noble Gas. Na is an alkali metal. F is halogen.
Group 17 is halogens. Inert is Noble Gases. Odourless and colourless is Noble Gases. Alkali metals do not occur freely in nature. Alkali metals are malleable
Explanation:
the second option is the correct answer i PROMISE
Answer:
Mass = 58.96 g
Explanation:
Given data:
Mass of CH₄ = 21.5 g
Mass of O₂ = 387.5 g
Mass of CO₂ formed = ?
Solution:
Chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
Number of moles of CH₄:
Number of moles = mass / molar mass
Number of moles = 21.5 g/ 16 g/mol
Number of moles = 1.34 mol
Number of moles of O₂ :
Number of moles = mass / molar mass
Number of moles = 387.5 g/ 32 g/mol
Number of moles = 12.1 mol
now we will compare the moles of CO₂ with O₂ and CH₄.
O₂ : CO₂
2 : 1
12.1 : 1/2×12.1 = 6.05 mol
CH₄ : CO₂
1 : 1
1.34 : 1.34
Number of moles of CO₂ produced by CH₄ are less thus it will limiting reactant.
Mass of CO₂:
Mass = number of moles × molar mass
Mass = 1.34 mol × 44 g/mol
Mass = 58.96 g
