Answer:
-290KJ/mol
Explanation:
ΔHrxn = ΔHproduct - ΔHreactant
ΔHrxn= 4ΔHH3PO4 - {6ΔHH2O + ΔHP4O10}
ΔHrxn = 4(-1279) - [6(-286) - 3110]
= -5116 -(-1716-3110)
= -5116-(-4826)
= -5116 + 4826 = -290KJ/mol
Answer:
3
Explanation:
the three elements involved in this compound are Li, S, O.
lithium, sulfur, and oxygen. Which create the ionic compound, "Lithium Sulfate."
7 atoms total, since there are two lithium, four oxygen, and one sulfate atom. this is a white inorganic salt.
Explanation:
To answer this question, we'll need to use the Ideal Gas Law:
p
V
=
n
R
T
,
where
p
is pressure,
V
is volume,
n
is the number of moles
R
is the gas constant, and
T
is temperature in Kelvin.
The question already gives us the values for
p
and
T
, because helium is at STP. This means that temperature is
273.15 K
and pressure is
1 atm
.
We also already know the gas constant. In our case, we'll use the value of
0.08206 L atm/K mol
since these units fit the units of our given values the best.
We can find the value for
n
by dividing the mass of helium gas by its molar mass:
n
=
number of moles
=
mass of sample
molar mass
=
6.00 g
4.00 g/mol
=
1.50 mol
Now, we can just plug all of these values in and solve for
V
:
p
V
=
n
R
T
V
=
n
R
T
p
=
1.50 mol
×
0.08206 L atm/K mol
×
273.15 K
1 atm
= 33.6 L
this is not the answer but it will help you
do by the formula it is on the answer
Answer:
Explanation:
Hello!
In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:
We plug in the mass of water, temperature change and specific heat to obtain:
Now, this enthalpy of reaction corresponds to the combustion of propyne:
Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:
However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:
Now, we solve for the enthalpy of formation of C3H4 as shown below:
So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):
Best regards!
