Answer: The new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.
Explanation:
Given: 
= 61 L, 
= 183 K, 
= 0.60 atm
At STP, the value of pressure is 1 atm and temperature is 273.15 K.
Now, formula used to calculate the new volume is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that the new volume of a 61 L sample at STP that is moved to 183 K and 0.60 atm is 54.63 L.
It produces nitrogen gas and water
4NH3+3O2+heat 2N2+6H 2O
The dissociation equation will be
NH4OH ---> NH4+ + OH-
Initial 0.006 0 0
Change -0.006 X 0.053 +0.006 X 0.053 -0.006 X 0.053
Equlibrium 0.006 -0.006 X 0.053 0.006 X 0.053 0.006 X 0.053
Ka = [NH4+] [ OH-] / [NH4OH] = (0.006 X 0.053)^2 / 0.006 -0.006 X 0.053
Ka = 1.78 X 10^-5
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
A. The heat is needed to melt 100.0 grams of ice that is already at 0°C is +33,400 J.
<h3>What is Specific heat capacity?</h3>
Specific heat capacity is the quantity of heat needed to raise the temperature per unit mass.
<h3>
Heat needed to melt the cube of ice</h3>
The heat is needed to melt 100.0 grams of ice that is already at 0°C is calculated as follows;
Q = mL
where;
- m is mass of the ice
- L is latent heat of fusion of ice = 334 J/g
Q = 100 x 334
Q = 33,400 J
Thus, the heat is needed to melt 100.0 grams of ice that is already at 0°C is +33,400 J.
Learn more about heat capacity here: brainly.com/question/16559442
#SPJ1
