Answer:
The answer is
<h2>2 cm/year</h2>
Explanation:
To find the rate in cm/year we must first convert 200 m into cm
1 m = 100 cm
if 1 m = 100 cm
Then 200 m = 200 × 100 = 20 ,000 cm
So the rate is
<h2>
</h2>
<u>Reduce the fraction with 10,000</u>
We have the final answer as
<h3>2 cm/year</h3>
Hope this helps you
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
Answer:
The number of energy levels will increase.
Explanation:
As they are all Noble Gases, they are all in the same family. However, as you go further down the list of Noble Gases, the period number increases. The period number shows the number of energy levels. Hence, an increase in energy levels.
Explanation:
<h3>The mixture in which the particles of the components of solute and solvent are equally mixed is called homogeneous mixture. </h3>
Combustion equation of n-hexane:
2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O
a)
Assuming we have 100 moles of air,
Oxygen = 20.9 moles
n-hexane required = 20.9/19 x 2
= 2.2 moles
LFL = Half of stoichometric amount = 2.2 / 2 = 1.1
LFL n-hexane = 1.1%
b)
1.1 volume percent required for LFL
1.1% x 1
= 0.0011 m³ of n-hexane required
