Answer:
The mass of FeSO4*7H2O in the sample is 1.21 grams
Explanation:
<u>Step 1</u>: Calculate moles of Fe2O3
moles of Fe2O3 = mass of Fe2O3 / Molar mass of Fe2O3
moles of Fe2O3 = 0.348 grams / 159.69 g/mole = 0.00218 moles
<u>Step 2</u>: Calculate moles of Fe
4 Fe + 3O2 → 2Fe2O3
For 4 moles of Fe consumed there is 2 moles of Fe2O3 produced
This means it has a ratio 2:1
So 0.00218 moles of Fe2O3 produced , there is 2*0.00218 = 0.00436 moles of Fe consumed
<u>Step 3:</u> Calculate moles of FeSO4*7H2O
Fe + H2SO4 + 7H2O → FeSO4*7H20 + H2
For 1 mole of Fe consumed there is 1 mole of FeSO4*7H2O produced
This means for 0.00436 moles there is 0.00436 moles of Fe2SO4*H2O produced
<u>Step 4:</u> Calculate the mass of FeSO4*7H2O in the sample
mass of FeSO4*7H2O = 0.00436 moles * 278.01 g/mole = 1.212 g
The mass of FeSO4*7H2O in the sample is 1.21 grams
Explanation:
The given data is as follows.
,
= 2257 kJ/kg,
= ?
For water,
= 4.184 
Formula to calculate heat of vaporization is as follows.
= 
Hence, putting the values into the above formula as follows.
= 
= 
= 2257 kJ/kg - 376.56 kJ/kg
= 1880.44 kJ/kg
Thus, we can conclude that enthalpy of liquid water at
is 1880.44 kJ/kg.
Once for the water and once for the copper. Set up a table that accounts for each of the variables you know, and then identify the ones you need to obtain. Give me a moment or two and I will work this out for you.
Okay, so like I said before, you will need to use the equation twice. Now, keep in mind that when the copper is placed in the water (the hot into the cold), there is a transfer of heat. This heat transfer is measured in Joules (J). So, the energy that the water gains is the same energy that the copper loses. This means that for your two equations, they can be set equal to each other, but the copper equation will have a negative sign in front to account for the energy it's losing to the water.
When set equal to each other, the equations should resemble something like this:
(cmΔt)H20 = -(cmΔt)Cu
(Cu is copper).
Remember, Δt is the final temperature minus the initial temperature (T2-T1). We are trying to find T2. Since we are submerging the copper into the water, we can assume that the final temperature at equilibrium is the same for both the copper and the water. At a thermodynamic equilibrium, there is no heat transfer because both materials are at the same temperature.
T2Cu = T2H20
Now, the algebra for this part of the problem is a bit confusing, so make sure you keep track of your variables. If done right, the algebra should work out so you have this:
T2 = ((cmT1)Cu + (cmT1)H20) / ((cm)H20 + (cm)Cu)
Insert the values for the variables. Once you plug and chug, your final answer should be
26.8 degrees Celsius.
Answer:
2.5 moles of NaCl
Explanation:
The balanced chemical reaction equation is shown in the image. Since it takes 2 moles of Hydrochloric acid to form two moles of sodium. Chloride, then 2.5 moles of hydrochloric acid should also form 2.5 moles of sodium chloride according to the balanced reaction equation.
Answer:
MgSO4.7H2O
Explanation:
Let the formula for the hydrated magnesium sulphate be MgSO4.xH2O
Mass of the hydrated salt (MgSO4.xH2O) = 12.845g
Mass of anhydrous salt (MgSO4) = 6.273g
Mass of water molecule(xH2O) = Mass of the hydrated salt — Mass of anhydrous salt = 12.845 — 6.273 = 6.572g
Now,we can obtain the number of mole of water molecule present in the hydrated salt as follows:
Molar Mass of hydrated salt (MgSO4.xH2O) = 24 + 32 + (16x4) + x(2 + 16) = 24 + 32 + 64 + x(18) = 120 + 18x
Mass of xH2O/ Molar Mass of MgSO4.xH2O = Mass of water / mass of hydrated salt
18x/120 + 18x = 6.572/12.845
Cross multiply to express in linear form
18x x 12.845 = 6.572(120 + 18x)
231.21x = 788.64 + 118.296x
Collect like terms
231.21x — 118.296x = 788.64
112.914x = 788.64
Divide both side by 112.914
x = 788.64 /112.914
x = 7
Therefore the formula for the hydrated salt (MgSO4.xH2O) is MgSO4.7H2O