Answer:
44.2 L
Explanation:
Use Charles Law:

We have all the values except for V₂; this is what we're solving for. Input the values:
- make sure that your temperature is in Kelvin
From here, we need to get V₂ by itself. To do this, multiply by 273 on both sides:

Therefore, V₂ = 44.2 L
It's also helpful to know that temperature and volume are linearly related. So, when temperature drops, so will volume and vice versa.
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:
Bases do not react with metals in the way that acids do
Explanation:
hope this helps
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Answer:
28.75211 kj
Explanation:
Given data:
Mass of iron bar = 841 g
Initial temperature = 84°C
Final temperature = 7°C
Heat released = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
specific heat capacity of iron is 0.444 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 7°C - 84°C
ΔT = -77°C
By putting values,
Q = 841 g × 0.444 j/g.°C × -77°C
Q = 28752.11 j
In Kj:
28752.11 j × 1 kJ / 1000 J
28.75211 kj