Answer:
1120 gm
Explanation:
6. Consider the reaction: CzHo (g) + 02 (8) - 4 CO2(g) + 6H2O (1)
(a) Balance the equation.
(b) How many grams of oxygen are required to react with 10 moles of ethane for a complete
combustion reaction?
FIRST, CORRECT THE EQUATION THEN BALANCE
2C2H6(G) + 7O2------------> 4CO2 + 6H2O
so for 10 moles of ethane, we need
7 X 5 = 35 MOLES O2
=35 MOLES O2
O2 HAS A MOLAR MASS OF 2X16 = 32 gm
35 MOLES OF O2 HAS A MASS OF 35 X 32 =1120 gm
Mass atomic of Ne=20.18 u
Therefore:
molar mass=20.18 g/1 mol
1 mole=6.022*10²³ particles (atoms or molecules)
Then: 6.022*10²³ atoms are contained in 20.18g
Now, We can solve this problem by the three rule.
6.022*10²³ atoms-------------------20.18 g
x------------------------------------------32 g
x=(6.022*10²³ atoms * 32 g)/20.18 g=9.55*10²³ atoms.
Answer: 9.55*10²³ Ne atoms are contained in 32 g of the element.
<span>C) The element nitrogen is made up of carbon and nitrogen</span>
Answer:
Explanation:
Question 7.
We can use the Combined Gas Laws to solve this question.
a) Data
p₁ = 1.88 atm; p₂ = 2.50 atm
V₁ = 285 mL; V₂ = 435 mL
T₁ = 355 K; T₂ = ?
b) Calculation
Question 8. I
We can use the Ideal Gas Law to solve this question.
pV = nRT
n = m/M
pV = (m/M)RT = mRT/M
a) Data:
p = 4.58 atm
V = 13.0 L
R = 0.082 06 L·atm·K⁻¹mol⁻¹
T = 385 K
M = 46.01 g/mol
(b) Calculation
Answer: Once the fingerprints are captured and described, judicial physicists use a categorization structure to label the ruling class. The three elementary mark on finger patterns is Whorl, Arch, and Loop. There are more intricate categorization methods that further decay the pattern to plain arches or make camp arches. Loops grant permission be branching or ulnar.
