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romanna [79]
4 years ago
3

which is required for convection to occur? a.a solid substance b.a heat source c.boiling water d.air molecules

Physics
2 answers:
masha68 [24]4 years ago
5 0

Answer:

b.a heat source

Explanation:

Convection occurs in fluids. This phenomenon occurs when a fluid is heated: a classical example is the pot containing water over a flame. As the water is heated by the flame, the bottom part of the water becomes warmer than the upper part: as a result, the bottom part of the water goes up, replacing the colder part which goes down. Then, this colder part is heated, and it goes up again, etc. in a cycle. This phenomenon cannot occur if there is no heat source that heats the fluid.

Serjik [45]4 years ago
3 0
The answer is ( B. a heat  source ) Thank you have a nice day 

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enyata [817]

Answer:

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3 years ago
Read 2 more answers
Absolute zero (K=0 or -273.15°C) is what temperature on the Farenheit scale? a) 459.4°F b) -301.43 °F c) 233.05°F d) -40.15°F e
ANEK [815]

Answer:

e) None of these is true

Explanation:

Given that

Temperature = 0 K

We know that relationship between kelvin and Farenheit scale

\dfrac{K-273}{100}=\dfrac{F-32}{180}

Now by putting the values

\dfrac{K-273}{100}=\dfrac{F-32}{180}

\dfrac{0-273}{100}=\dfrac{F-32}{180}

So F= - 459.67°F

So we can say that 0 K is equal to  - 459.67°F.

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3 0
4 years ago
Two molecules of lithium are combined with 225 grams of bromine to form two molecules of lithium bromide. If you end up with 690
4vir4ik [10]

The amount of Li present to start the reaction is 55.18g

<u>Explanation:</u>

2Li + Br₂ → 2LiBr

Molecular weight of Br₂ = 159.808 g/mol

Mass of Br₂ present = 225 g

Moles of Br₂ present during the reaction = 225 / 159.808

                                                                m = 1.4

Molecular weight of LiBr = 86.845 g/mol

Mass of LiBr formed = 690 g

moles of LiBr produced = 690 / 86.845

                               m(LiBr) = 7.95

According to the balanced equation, 2 molecules of Li reacts to for 2 molecule of LiBr

So, 7.95 moles of LiBr would require 7.95 moles of Li

The molecular weight of Li is 6.941 g/mol

Thus, the amount of lithium present to start the reaction is

moles = \frac{given weight}{molecular weight} \\\\7.95 = \frac{w}{6.941} \\\\w = 55.18g

Therefore, the amount of Li present to start the reaction is 55.18g

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Answer:

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