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Fed [463]
2 years ago
10

Why using a temperature probe is preferable to using a conventional mercury thermometers

Chemistry
1 answer:
Crank2 years ago
3 0
This for health and hazard purposes. The conventional thermometers contain mercury which is very dangerous to health. It is toxic which could cause poisoning, respiratory damage and kidney problems. Because of this, it is already illegal to use them for safety purposes. Temperature probes are much safer because they only use thermocouples.
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List 3 scenarios for a substance being classified as a weak acid
malfutka [58]
A way to classify an acid as being weak is to determine how strongly the substance conducts electricity. If the acid or base conducts electricity strongly, it is a strong acid or base. If the acid or base conducts electricity weakly, it is a weak acid or base.
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3 years ago
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A celestial body is any natural (i.
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What type(s) of bond(s) does carbon have a tendency to form?
Naily [24]
Carbon has a tendency to form covalent bonds.
4 0
3 years ago
2. What is the mass of a car with a momentum of 10,000.00 kgm/s, traveling at 15.0 m/s?
Svetradugi [14.3K]

Answer:

666.67 kg is the mass of a car.

Explanation:

Momentum is defined as amount of motion possessed by the the moving body. It is mathematically calculated by multiplying mass into velocity by which object is moving.

Momentum(P)=Mass(m)\times Velocity(v)

Mass of the car = m =?

Velocity of the car = v = 15.0 m/s

Momentum of the car = P = 10,000.00 kgm/s

m=\frac{P}{v}

m=\frac{ 10,000.00 kgm/s}{15.0 m/s}=666.67 kg

666.67 kg is the mass of a car.

7 0
2 years ago
Boyles Law P1V1 = P2V2
arsen [322]

Answer:

A. The balloons will increase to twice their original volume.

Explanation:

Boyle's law states that the pressure exerted on a gas is inversely proportional to the volume occupied by the gas at constant temperature. That is:

P ∝ 1/V

P = k/V

PV = k (constant)

P = pressure, V = volume.

P_1V_1=P_2V_2

Let the initial pressure of the balloon be P, i.e. P_1=P, initial volume be V, i.e. V_1=V. The pressure is then halved, i.e. P_2=\frac{P}{2}

P_1V_1=P_2V_2\\\\P*V=\frac{P}{2} *V_2\\\\V_2=\frac{2*P*V}{P}\\\\V_2=2V

Therefore the balloon volume will increase to twice their original volume.

3 0
2 years ago
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