Firstly the limiting reactant should be identified. Limiting reactant is the reactant that is in limited supply, the amount of product formed depends on the moles present of the limiting reactant.
the stoichiometry of x to y = 1:2
1 mole of x reacts with 2 moles of y
if x is the limiting reactant, there are 3 moles of x, then 6 moles of y should react, however there are only 4 moles of y. Therefore y is the limiting reactant and x is in excess.
4 moles of y reacts with 2 moles of x
since there are 3 moles of x initially and only 2 moles are used up, excess amount of x is 1 mol thats in excess.
Answer:
TATGGCGTT
Explanation:
Complimentary base pairs:
A-T
C-G
Use the other letter for complimentary strands
Answer:
Mass = 0.697 g
Explanation:
Given data:
Volume of hydrogen = 1.36 L
Mass of ammonia produced = ?
Temperature = standard = 273.15 K
Pressure = standard = 1 atm
Solution:
Chemical equation:
3H₂ + N₂ → 2NH₃
First of all we will calculate the number of moles of hydrogen:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
1atm ×1.36 L = n × 0.0821 atm.L/mol.K × 273.15 K
1.36 atm.L = n × 22.43 atm.L/mol
n = 1.36 atm.L / 22.43 atm.L/mol
n = 0.061 mol
Now we will compare the moles of hydrogen and ammonia:
H₂ : NH₃
3 : 2
0.061 : 2/3×0.061 = 0.041
Mass of ammonia:
Mass = number of moles × molar mass
Mass = 0.041 mol × 17 g/mol
Mass = 0.697 g
Jill and Susan violated safety procedures by not properly listening and/or reading over the instructions to know all the materials, steps, and equipment they need for the lab. Hope this helps!