Answer:
116.3 grCO2
Explanation:
1st - we balance the equation so that it finds the same amount of elements of the product side and of the reagent side
C6H6 +15/2 O2⟶ 6CO2 +3 H2O
2nd - we calculate the limiting reagent
39.2gr C6H6*(240grO2/78grC6H6)=120 grO2
we don't have that amount of oxygen so this is the excess reagent and oxygen the limiting reagent
3rd - we use the limiting reagent to calculate the amount of CO2 in grams
105.7grO2*(264grCO2/240grO2)=116.3 grCO2
We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+ + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:
PH= -㏒[H+]
= -㏒(4.81x10^-7) = 6.32
Answer:
d is your answer please follow me
Answer:
0.719M AgNO₃
Explanation:
Based on the reaction:
MgBr₂ + 2AgNO₃ ⇄ 2AgBr + Mg(NO₃)₂
<em>1 mole of magnesium bromide reacts completely with 2 moles of AgNO₃</em>
<em />
To find molarity of AgNO₃ solution we need to determine moles of AgNO₃ and, as molarity is the ratio of moles over liter (13.9mL = 0.0139L). Now, to determine moles of AgNO₃ we need to use the reaction, thus:
<em>Moles AgNO₃:</em>
<em />
Moles of MgBr₂ are:
50.0mL = 0.050L * (0.100mol / L) = 0.00500 moles of MgBr₂.
As the silver nitrate reacts completely and 2 moles of AgNO₃ reacts per mole of MgBr₂:
0.00500 moles MgBr₂ * (2 moles AgNO₃ / 1 mole MgBr₂) =
0.0100 moles of AgNO₃ are in the solution.
And molarity is:
0.0100 moles AgNO₃ / 0.0139L =
<h3>0.719M AgNO₃</h3>
Answer:
<h2>119.60 moles</h2>
Explanation:
To find the number of moles in a substance given it's number of entities we use the formula
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
We have the final answer as
<h3>119.60 moles</h3>
Hope this helps you
