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2 years ago

Differentiate the three pictures and explain them briefly.

Chemistry

1 answer:

Dmitry [639]2 years ago

8 0

Answer:

answer

Cold Front - a zone separating two air masses, of which the cooler, denser mass is advancing and replacing the warmer.

Warm Front - a transition zone between a mass of warm air and the cold air it is replacing.

Stationary Front - a front between warm and cold air masses that is moving very slowly or not at all.

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If a drop of blood is 0.05 mL, how many drops of blood are in a blood collection tube that holds 2 mL ?

soldi70 [24.7K]

Just divide the two (2 / 0.05) and you will get your answer; there are 40 drops of bloodin the collection tube.

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3 years ago

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A point of reference is anything that seems steady that is used compare the position of an object

Masja [62]

Answer:

True

Explanation:

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2 years ago

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what temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution (specific heat capacity = 3

ivolga24 [154]

Answer:

1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.

Explanation:

Mass of ethylene glycol = m = 100 g

Specific heat capacity of ethylene glycol = c = 3.5 J/g°C

Change in temperature of ethylene glycol = ΔT

Heat loss by the ethylene glycol = Q = 350 J

$Q=mc\Delta T$

$\Delta T=\frac{Q}{mc}=\frac{350 J}{100 g\times 3.5 J/g^oC}$

ΔT = 1°C

1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.

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3 years ago

Question 10

olga2289 [7]

Answer:

Here it would be A,B and C because all the examples are violating safety precautions. They are all examples of un safe signs.

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Explanation:

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2 years ago

Iodine chloride, ICl, can be made by the following reaction between iodine, I2, potassium iodate, KIO3, and hydrochloric acid.

Vadim26 [7]

The balanced chemical reaction is:

2 I2 + KIO3 + 6 HCl ---------> 5 ICl + KCl + 3 H2O

We are given the amount of the product to be produced from the reaction. This will be the starting point of our calculations.

28.6 g ICl (1 mol / 162.35 g ICl ) ( 2 mol I2 / 5 mol ICl ) ( 253.81 g I2 / 1 mol I2 ) = 17.88 g I2

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3 years ago

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