Answer:
I say for her to test her suspicion by going to see an allergist to diagnose the allergy.
The answer is atoms good luck
Answer:
Oxidation–reduction or redox reactions are reactions that involve the transfer of electrons between chemical species (check out this article on redox reactions if you want a refresher!). The equations for oxidation-reduction reactions must be balanced for both mass and charge, which can make them challenging to balance by inspection alone. In this article, we’ll learn about the half-reaction method of balancing, a helpful procedure for balancing the equations of redox reactions occurring in aqueous solution.
Explanation:
Answer:a) 11.34 g of ethane 
can be formed
b) 
is the limiting reagent
c) 3.44 g of the excess reagent remains after the reaction is complete
Explanation:
To calculate the moles :
1. 
2. 
According to stoichiometry :
1 mole of require 1 mole of 
Thus 0.378 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
moles of left = (2.10-0.378) = 1.72 moles
mass of left=
According to stoichiometry :
As 1 mole of give = 1 mole of 
Thus 0.378 moles of give = of
Mass of 
Thus 11.34 g of ethane is formed.
volume of H₂O = 7.2 L
Explanation:
The combustion reaction of methane (CH₄):
CH₄ + 2 O₂ → CO₂ + 2 H₂O
Now we calculate the number of moles of methane using the following formula:
number of moles = volume / 22.4 (L/mole)
number of moles of CH₄ = 3.6 / 22.4
number of moles of CH₄ = 0.16 moles
Taking in account the chemical reaction, we devise the following reasoning:
if 1 mole of CH₄ produce 2 moles of H₂O
then 0.16 moles of CH₄ produce X moles of H₂O
X = (0.16 × 2) / 1 = 0.32 moles of H₂O
And now we can calculate the volume of water (H₂O) produced by the reaction:
number of moles = volume / 22.4 (L/mole)
volume = number of moles × 22.4 (L/mole)
volume of H₂O = 0.32 × 22.4
volume of H₂O = 7.2 L
Learn more about:
combustion reaction
brainly.com/question/14122510
#learnwithBrainly
