The answer would be C.
The salt and the water have both undergone physical changes.
Hope this helps!
The number of grams of Ag2SO4 that could be formed is 31.8 grams
<u><em> calculation</em></u>
Balanced equation is as below
2 AgNO3 (aq) + H2SO4(aq) → Ag2SO4 (s) +2 HNO3 (aq)
- Find the moles of each reactant by use of mole= mass/molar mass formula
that is moles of AgNO3= 34.7 g / 169.87 g/mol= 0.204 moles
moles of H2SO4 = 28.6 g/98 g/mol =0.292 moles
- use the mole ratio to determine the moles of Ag2SO4
that is;
- the mole ratio of AgNo3 : Ag2SO4 is 2:1 therefore the moles of Ag2SO4= 0.204 x1/2=0.102 moles
- The moles ratio of H2SO4 : Ag2SO4 is 1:1 therefore the moles of Ag2SO4 = 0.292 moles
- AgNO3 is the limiting reagent therefore the moles of Ag2SO4 = 0.102 moles
<h3> finally find the mass of Ag2SO4 by use of mass=mole x molar mass formula</h3>
that is 0.102 moles x 311.8 g/mol= 31.8 grams
Balanced equation: shown in photo
What class is this for because it depends
Answer:
1,085g of water
Explanation:
If we have the value 4520kj is because the question is related to Energy and heat capacity. In this case, the law and equation that we use is the following:
Q= m*C*Δt where;
Q in the heat, in this case: 4520kj
m is the mas
Δt= is the difference between final-initial temperature (change of temperature), in this exercise we don´t have temperatura change.
In order to determine the mass, I will have the same equation but finding m
m= Q/C*Δt without m=Q/C
So: m= 4,520J/4.18J/g°C
m= 1,0813 g
