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nydimaria [60]
3 years ago
9

The molecules which evaporate into the gas phase are the ones traveling with ________ velocity.

Chemistry
2 answers:
sveticcg [70]3 years ago
5 0
The appropriate response is high. The best approach to consider it is those molecules have more vitality than the normal and it is unequivocally those higher enthusiastic ones that have enough vitality to break the intra-molecular forces and leave. By the way this is the reason when we sweat we chill off - the most enthusiastic molecules vanish diverting warmth.
yarga [219]3 years ago
3 0

Explanation:

In gases, the molecules are held by weak Vander waal forces. Hence, they have high kinetic energy due to which they tend to move rapidly from one place to another leading to more number of collisions.  

Therefore, when number of collisions increases then force applied by the molecules on each other also increases.

As a result, velocity of the particles increases.

Therefore, we can conclude that the molecules which evaporate into the gas phase are the ones traveling with high velocity.

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Which general equation shows a Single-displacement reaction?
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6 0
2 years ago
Enter a molecular equation for the gas-evolution reaction that occurs when aqueous hydroiodic acid and aqueous potassium sulfite
Katen [24]

Answer:

2HI + K2SO3=>2KI+H2SO3

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3 years ago
A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
mamaluj [8]

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

Molarity=\frac{moles}{\text{Volume of solution(L)}}

Moles of glucose = \frac{18.5 g}{180 g/mol}=0.1028 mol

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = \frac{0.1028 mol}{0.1 L}=1.028 mol/L

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

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Volume of the solution taken = V_1=30.0 mL

Molarity of the solution after dilution = M_2

Volume of the solution after dilution= V_2=0.500L = 500 mL

M_1V_1=M_2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}

M_2=0.07248 mol/L

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}

Moles of glucose = 0.07248 mol/L\times 0.1 L=0.007248 mol

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

4 0
4 years ago
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