Step 7- Communicate. Present/share your results. Replicate.
Step 1- Question.
Step 2-Research.
Step 3-Hypothesis.
Step 4-Experiment.
Step 5-Observations.
Step 6-Results/Conclusion.
Answer:
d) V = 91.3 L
Explanation:
Given data:
Volume of nitrogen = ?
Temperature = standard = 273.15 K
Pressure = standard = 1 atm
Number of atoms of nitrogen = 2.454×10²⁴ atoms
Solution:
First of all we will calculate the number of moles of nitrogen by using Avogadro number.
1 mole = 6.022×10²³ atoms
2.454×10²⁴ atoms × 1 mol / 6.022×10²³ atoms
0.407×10¹ mol
4.07 mol
Volume of nitrogen:
PV = nRT
1 atm × V = 4.07 mol ×0.0821 atm.L /mol.K ×273.15 K
V = 91.3 atm.L /1 atm
V = 91.3 L
Answer:
Molecular formula = C20H30
Explanation:
NB 440mg = 0.44g, 135mg= 0.135g
From the question, moles of CO2= 0.44/44= 0.01mol
Since 1 mol of CO2 contains 1mol of C, it implies mol of C = 0.01
Also from the question, moles of H2O = 0.135/18= 0.0075mole
Since 1 mol of H2O contains 2mol of H, it implies mol of H = 0.0075×2= 0.015 mol of H
To get the empirical formula, divide by smallest number of mole
Mol of C = 0.01/0.01=1
Mol of H = 0.015/0.01= 1.5
Multiply both by 2 to obtain a whole number
Mol of C =1×2 = 2
Mol of H= 1.5×2 = 3
Empirical formula= C2H3
[C2H3] not = 270
[ (2×12) + 3]n = 270
27n = 270
n=10
Molecular formula= [C2H3]10= C20H30
<span>J.j thomson a british physicist was the first to identify the electron in 1987</span>
