A. I think sorry if it’s wrong
¹/3 C3H8(g) + ⁵/3 O2(g)
Explanation:
The coefficient before every molecule is representative of the number of moles. We can represent it in ration form so as to calculate the question;
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l) means;
For every 1 mole of C₃H₈(g) and 5 moles of O₂(g) produces 3 moles of CO₂(g) and 4 moles of H₂O(l).
Therefore to produce 1.00 mole of CO₂(g);
We represent it in ratio;
C₃H₈(g) : CO₂(g)
1 : 3
What about ;
? (x) : 1
We cross multiply;
3x = 1 * 1
X = 1/3
We evaluate the same for O₂;
O₂(g) : CO₂(g)
5 : 3
What about
? (x) : 1
3x = 5 * 1
x = 5/3
Learn More:
For more on evaluating moles in chemical reactions check out;
brainly.com/question/13967925
brainly.com/question/13969737
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Answer is: A) Sr (strontium).
The reactivity series<span> is a series of metals from highest to lowest reactivity.</span><span> Metal higher in the reactivity series will displace another.
</span>Strontium is only higher in this group from magnesium. Strontium is stronger reducing agent than magnesium, gives electrons easier.
Answer:
Carbon dioxide and hydrogen monoxide
CaCl2 and KCl are both salts which dissociate in water
when dissolved. Assuming that the dissolution of the two salts are 100 percent,
the half reactions are:
<span>CaCl2 ---> Ca2+ + 2 Cl-</span>
KCl ---> K+ + Cl-
Therefore the total Cl- ion concentration would be coming
from both salts. First, we calculate the Cl- from each salt by using stoichiometric
ratio:
Cl- from CaCl2 = (0.2 moles CaCl2/ L) (0.25 L) (2 moles
Cl / 1 mole CaCl2)
Cl- from CaCl2 = 0.1 moles
Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1
mole KCl)
Cl- from KCl = 0.1 moles
Therefore the final concentration of Cl- in the solution
mixture is:
Cl- = (0.1 moles + 0.1 moles) / (0.25 L + 0.25 L)
Cl- = 0.2 moles / 0.5 moles
<span>Cl- = 0.4 moles (ANSWER)</span>
