Molar mass of N = 14 g/mol
Molar mass of O2 = 32 g/mol
Adding both masses = 46 g/mol
Actual molar mass/ Empirical molar mass = 138.02 / 46 = 3
Now multiplying this co effecient with empirical fomula NO2 = 3(NO2)
= N3O6
So according to above explanation,
<span>D) N3O6, is the correct answer.</span>
Answer:

Explanation:
Given that,
Heat absorbed by a sample, Q = 228 J
Mass of a sample, m = 706 g = 0.706 kg
Initial temperature is 26 °C and final temperature is 88.8°C
We need to find the heat absorbed by the sample. The heat absorbed by an object is given by :

So, the specific heat of the sample is
.
Answer:
The change in T° is 47.1 °C
Explanation:
Calorimetry formula to solve this:
Q = m . C . ΔT
We replace the data gien:
635 J = 3.50×10⁻² kg . 0.385 J/g°C . ΔT
In units of C, we have g and the mass (m) is in kg. So let's convert it from kg to g → 3.50×10⁻² kg . 1000 g / 1kg = 35 g. Now we can determine the ΔT:
635 J = 35 g . 0.385 J/g°C . ΔT
635 J / 35 g . 0.385 J/g°C = ΔT
47.1°C = ΔT
<span> -101.5 celcius and exactly that
</span>
Answer:
The correct approach is "12.25°C".
Explanation:
Given:
Mass of lead,
mc = 245 g
Initial temperature,
tc = 300°C
Mass of Aluminum,
ma = 150 g
Initial temperature,
ta = 12.0°C
Mass of water,
mw = 820 g
Initial temperature,
tw = 12.0°C
Now,
The heat received in equivalent to heat given by copper.
The quantity of heat = 
then,
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 