The heat that creates this temperature change coming from change in the internal energy of the system as per as first law of thermodynamics.
<h3>What is Boyle's law ?</h3>
A law stating that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.
As we know, Boyle's law only works when the gas is kept at a constant temperature
Here,
When volume of gases decreased, it means work done has occurred on the system, so the work done is used for raising internal energy of the gas and the other is released as the thermal energy.
So,
According to 1st law of thermodynamics,
we know Q = ΔU + W i.e, change in internal energy and work done. So this is a reason. Changing temperature occurs.
Learn more about Internal enrgy here ;
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<h3><u>Answer and explanation</u>;</h3>
- <em><u>The isotope U-235 is an important common nuclear fuel because under certain conditions it can readily be split, yielding a lot of energy. It is therefore said to be 'fissile' and use the expression 'nuclear fission'.</u></em>
- <em><u>Uranium 238 on the other hand is not fissionable by thermal neutrons, but it can undergo fission from fast or high energy neutrons. Hence it is not fissile, but it is fissionable.</u></em>
- In a nuclear power station fissioning of uranium atoms replaces the burning of coal or gas. Heat created by splitting the U-235 atoms is then used to make steam which spins a turbine to drive a generator, producing electricity.
Answer:
8.8g of Al are necessaries
Explanation:
Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.
To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:
<em>Moles H2:</em>
PV = nRT; PV/RT = n
<em>Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP</em>
Replacing:
1atm*11L/0.082atmL/molK*273.15K = n
n = 0.491 moles of H2 must be produced
<em />
<em>Moles Al:</em>
0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required
<em />
<em>Mass Al -Molar mass: 26.98g/mol-:</em>
0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries
Answer:
50
Explanation:
We will need a balanced equation with masses, moles, and molar masses of the compounds involved.
1. Gather all the information in one place with molar masses above the formulas and masses below them.
Mᵣ: 30.01 32.00 46.01
2NO + O₂ ⟶ 2NO₂
Mass/g: 80.00 16.00
2. Calculate the moles of each reactant

3. Calculate the moles of NO₂ we can obtain from each reactant
From NO:
The molar ratio is 2 mol NO₂:2 mol NO

From O₂:
The molar ratio is 2 mol NO₂:1 mol O₂

4. Identify the limiting and excess reactants
The limiting reactant is O₂ because it gives the smaller amount of NO₂.
The excess reactant is NO.
5. Mass of excess reactant
(a) Moles of NO reacted
The molar ratio is 2 mol NO:1 mol O₂

(b) Mass of NO reacted

(c) Mass of NO remaining
Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO