1) Recording the change in pressure when heating an inflated tire
Explanation:
This happens because with an increase in temperatures, the kinetic energy of the gas in the tire increases. The increase in kinetic energy means the molecules move more rapidly. There is an increased rate of collisions between the gas molecules, and increased energy of the collisions, and between the gas molecules and inner walls of the tire hence the increased pressure. Pressure and temperatures and directly proportional s long as the volume of the gas is kept constant.
Answer: 2
Explanation: Potassium, K, has only 1 valence electron, 4s^1. Oxygen, O, needs 2 more electrons to complete its valence shell to make 8 electrons. That means 2 K atoms will combine with 1 O atom, to produce K2O, potassium oxide. Both K electrons are stolen by the single O atom, so 2 electrons are transferred.
The thing that governs whether a reaction is exothermic is the energy given out / used up to break / form the bonds in the reaction.
<span>When two substances react, the bonds in those substances first break up, releasing energy, before re-forming in a different way, taking in energy. The nature of the bonds that are broken up and reformed determines whether more energy is given out (exothermic) or taken in (endothermic)</span>
Answer:
When a cold front approaches there can be a sudden drop in temperature, strong wind gusts, and heavy rain and thunderstorms.
Explanation:
The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
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