Answer:
V = 42.6 L
Explanation:
Given data:
Number of moles of Cl₂ = 1.9 mol
Temperature and pressure = standard
Volume occupy = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
By putting values,
1 atm × V = 1.9 mol ×0.0821 atm.L /mol.K × 273.15 k
V = 42.6 atm.L / 1 atm
V = 42.6 L
Answer:
<u>2.26 </u>
<u></u>
Explanation:
First you should know what are significant figures,
1 . All non - zero digits are significant.
2 . The zero between two non- zero are significant.Ex 302 has 3 significant figures.
3. The zero before the decimal and before any non-zero digit is non significant. Example : 0.003 has only 1 significant figure.
4. The zero after non zero are non - significant . But the zeros after the decimal point are significant.
300 has only 1 significant figure
300.0 has 4 significant figure
The least number of significant figures present in the number should be there in the final answer of the calculation.
Look at the number having minimum significant digits :
It is 3 (every number has 3- significant figures ) So the answer should also contain 3 - significant figures.

First solve the numerator part


Round off this number to 3 significant figures.
Answer is =<u> 2.26</u>
It has 3 - significant digit since all the digits are non-zero.
POH = - log [OH]
<span>pOH = - log (6.40 × 10-3) </span>
<span>pOH = 2.19 </span>
<span>pH + pOH = 14 </span>
<span>pH = 14 - pOH </span>
<span>pH = 14 - 2.19 </span>
<span>pH = 11.8</span>
Answer:
1. D (24.0 moles CO2)
2. A (.239 moles H2)
Explanation:
1. First Balance the equation
1 C3H8 + 5 O2 ---> 3 CO2 + 4 H2O
Then set up a stoiciometric equation so that the moles of O2 cancel out
40mol O2 x
= 24.0 moles CO2
2. Set up a stoichiometric equation
10 grams Fe x
x
= 0.239 moles H2