Answer:
220mol.
Explanation:
Water is H2O. Hydrogen gas is H2. Oxygen gas is O2. You have 220mol of O and 460mol of H. O is the limiting reactant. The ratio O:H2O is 1:1. 220*1=220
Answer:
86.3 g of N₂ are in the room
Explanation:
First of all we need the pressure from the N₂ in order to apply the Ideal Gases Law and determine, the moles of gas that are contained in the room.
We apply the mole fraction:
Mole fraction N₂ = N₂ pressure / Total pressure
0.78 . 1 atm = 0.78 atm → N₂ pressure
Room temperature → 20°C → 20°C + 273 = 293K
Let's replace data: 0.78 atm . 95L = n . 0.082 . 293K
(0.78 atm . 95L) /0.082 . 293K = n
3.08 moles = n
Let's convert the moles to mass → 3.08 mol . 28g /1mol = 86.3 g
Answer:
The answer to your question is C₂HO₃
Explanation:
Data
Hydrogen = 3.25%
Carbon = 19.36%
Oxygen = 77.39%
Process
1.- Write the percent as grams
Hydrogen = 3.25 g
Carbon = 19.36 g
Oxygen = 77.39 g
2.- Convert the grams to moles
1 g of H ----------------- 1 mol
3,25 g of H ------------- x
x = (3.25 x 1) / 1
x = 3.25 moles
12 g of C ---------------- 1 mol
19.36 g of C ---------- x
x = (19.36 x 1) / 12
x = 1.61 moles
16g of O --------------- 1 mol
77.39 g of O --------- x
x = (77.39 x 1)/16
x = 4.83
3.- Divide by the lowest number of moles
Carbon = 3.25/1.61 = 2
Hydrogen = 1.61/1.61 = 1
Oxygen = 4.83/1.61 = 3
4.- Write the empirical formula
C₂HO₃
Answer:
1.72 M
Explanation:
Molarity is the molar concentration of a solution. It can be calculated using the formula a follows:
Molarity = number of moles (n? ÷ volume (V)
According to the information provided in this question, the solution has 58.7 grams of MgCl2 in 359 ml of solution.
Using mole = mass/molar mass
Molar mass of MgCl2 = 24 + 35.5(2)
= 24 + 71
= 95g/mol
mole = 58.7g ÷ 95g/mol
mole = 0.618mol
Volume of solution = 359ml = 359/1000 = 0.359L
Molarity = 0.618mol ÷ 0.359L
Molarity = 1.72 M
D. As more electrons are added to an element, the number of electron orbitals being filled increases