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snow_lady [41]
3 years ago
13

explain placed a dented table -tennis ball in boiling water is one way to remove the dent in the ball assume the ball has no hol

es
Chemistry
1 answer:
goblinko [34]3 years ago
7 0
This is true because if you put the tennis ball in boiling water, oxygen molecules in the ball will move more vigorously, thus the dent will disappear.  
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When energy is transferred to ocean water from moving air above it what happens to most of the energy
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A 1.00 liter solution contains 0.24 M hypochlorous acid and 0.31 M sodium hypochlorite. If 0.160 moles of potassium hydroxide ar
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Explanation:

When OH- (as in potassium hydroxide) is added, it reacts with the acid (HOCl) to reduce the amount of HOCl and increase the concentration of  sodium hypochlorite.

Potassium hydroxide will react with the hypochlorous acid to produce hypochlorite ions. In the process, some of the weak acid will be consumed, along with the added strong base.

This occurs as follows:

HClO(aq) + KOH(aq) → KClO(aq) + H2O(l)  

since water is formed, this maintains the pH.  Thus ...

A. The number of moles of HClO will decrease. - TRUE

B. The number of moles of ClO- will increase. - TRUE

C. The equilibrium concentration of H3O+ will remain the same. - TRUE

D. The pH will decrease. - FALSE

E. The ratio of [HClO] / [ClO-] will decrease. -TRUE. It will decrease as HClO goes down and ClO- goes up.

5 0
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Which of the following is an electrolyte?<br> CH4OH<br> O Ca(NO3)2<br> O C12H22011<br> O CH3CH2OH
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Answer:

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Explanation:

5 0
3 years ago
Two bulbs are connected by a stopcock. The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 0.700 atm
ivolga24 [154]

Answer:

O_{2} and NO_{2}

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2.70 L*atm of nitric oxide reacts with (2.7/2) 1.35 L*atm of oxygen. This shows that there is more oxygen gas in the system than nitric oxide. Thus nitric oxide is the limiting reactant.

At the end of the experiment:

All the nitirc oxide has been used up, i.e. P_{NO} = 0

For the product: 2.70 L*atm NO produced  2.70 L*atm NO_{2}

The total volume of the system after the stopcock is opened = 6+1.5 = 7.5 L

The partial pressure of NO_{2}  = (2.70 L*atm NO_{2} ) / (7.5 L) = 0.36 atm NO_{2}  

Similarly for oxygen gas:

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Partial pressure of oxygen is:

2.40 L*atm / 7.5 L = 0.32 atm  

Thus, the gases present at the end of the experiment are O_{2} and NO_{2}

3 0
3 years ago
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