The diagram best shows the magnetic field around the bar magnet is the diagram in option B.
<h3>What is the magnetic field?</h3>
The magnetic field refers to the region in space where the influence of a magnet can be felt. We know that strength of magnet can be seen by following the magnetic lines of force. The more the magnetic lines of force, the stronger the magnetic field especially when the line are packed closely together.
Looking at the diagrams of the magnetic lines of force, it must be recalled that the magnetic lines of force runs from the north pole to the south pole and not the other way round.
Thus, the diagram best shows the magnetic field around the bar magnet is the diagram in option B.
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Answer:
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Explanation:
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moles
bonding
balancing equations
and set ups
Answer:
The new volume of the gas is 276.45 mL.
Explanation:
Charles's law indicates that for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases, and as the temperature decreases, the volume of the gas decreases.
Charles's law is a law that mathematically says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

Analyzing an initial state 1 and a final state 2, it is satisfied:

In this case:
- V1= 250 mL
- T1= 293 K
- V2= ?
- T2= 324 K
Replacing:

Solving:

V2= 276.45 mL
<em><u>The new volume of the gas is 276.45 mL.</u></em>
Answer:
The calorimeter constant is = 447 J/°C
Explanation:
The heat absorbed or released (Q) by water can be calculated with the following expression:
Q = c × m × ΔT
where,
c is the specific heat
m is the mass
ΔT is the change in temperature
The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.
The heat absorbed by the calorimeter (Q) can be calculated with the following expression:
Q = C × ΔT
where,
C is the calorimeter constant
The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).
Qabs + Qrel = 0
Qabs = - Qrel
Qcal + Qw₁ = - Qw₂
Qcal = - (Qw₂ + Qw₁)
Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)
Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) + (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]
Ccal = 447 J/°C
First you have to moles so multiply .0483L X .55M= .026565 Multiply moles by mole ratio which is 1/2, so the moles becomes .013283 now molarity=moles/volume; divide .013283/.015L=.885533M significant figures and you final answer is 0.89M