Answer:
= 4
Explanation:
To solve this exercise we will use Bohr's atomic model
= - 13.606 / n² [eV]
The transition from level n = 2 to level n = 1 is valid
= - 13.606 [¼ -1/1]
= 10.2045 eV
Bohr's model for atoms with only one electron is
= -13.606 Z² / n²
Where Z is the atomic number of the atom.
In this case the helium atom has an atomic number of Z = 2 from the level n₀ = 2 let's look up to what level it reaches
ΔE = -13.606 [4 / ² - 4/4]
4 / ² = -ΔE / 13.606 + 1
4 / ² = -10.2045 / 13.606 +1 = -0.75 +1
4 / ² = 0.25
= √ 4 / 0.25
= 4
We simply asked to name three uses for mercury.
The most common and well-known use of mercury is the production of thermometers. It's property to stay liquid at room temperature makes it ideal for a temperature indicator. However, the use of mercury is thermometers has been phased out due to health hazards.
It is also used to form an amalgam which is the result of its combination with silver or gold. Mercury has been used to mine gold and silver. This application has also been phased out.
Today's use of mercury includes mercury-vapor lamps which are the bright lamps used in high-ways.
Answer:
1. As rocket mass increases, acceleration decreases.
2. The inverse of the mass of the boat.
Explanation:
1. Newton's second law of motion states;
F = ma
where F is the force applied, m is the mass and a is the acceleration.
Therefore, increasing the mass of a rocket increases its weight which would reduce its acceleration provided that the force is constant. Thus, as rocket mass increases, acceleration decreases.
2. The slope of the graph can be expressed as;
From Newton's second law,
F = ma
Slope = (Δa) ÷ (ΔF)
Slope = 
⇒ 
=
Therefore, the slope of the graph is the reciprocal of the mass of the boat.
The specific heat of the unknown substance with a mass of 0.158kg is 0.5478 J/g°C
HOW TO CALCULATE SPECIFIC HEAT CAPACITY:
The specific heat capacity of a substance can be calculated using the following formula:
Q = m × c × ∆T
Where;
- Q = quantity of heat absorbed (J)
- c = specific heat capacity (4.18 J/g°C)
- m = mass of substance
- ∆T = change in temperature (°C)
According to this question, 2,510.0 J of heat is required to heat the 0.158kg substance from 32.0°C to 61.0°C. The specific heat capacity can be calculated:
2510 = 158 × c × (61°C - 32°C)
2510 = 4582c
c = 2510 ÷ 4582
c = 0.5478 J/g°C
Therefore, the specific heat capacity of the unknown substance that has a mass of 0.158 kg is 0.5478 J/g°C.
Learn more about specific heat capacity at: brainly.com/question/2530523
