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Travka [436]
3 years ago
14

when a metal, such as lead, is oxidied (loses electrons) to form a positive ion (cation), how does he solubility change?

Engineering
1 answer:
o-na [289]3 years ago
8 0

Answer: The size of the ion and the charge of the ion are the factors that affect solubility in water.

Explanation:

Lead lose electrons to become cations. Compounds with small ions tend to be less soluble than compounds with large ions. Large ions have higher solubility. This is because small ions are closely packed so it is difficult for water to break them apart.

Compounds with small ions seemingly have less solubility than those with large ions. The ions in the compound attract each other, and the water molecules attract the ions. Compounds would be soluble in water If the water molecules have a greater or higher attraction to the ions than ions have for each other.

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What is future active and future passive and future perfect active
san4es73 [151]
The future perfect tense forms are made by putting ‘will / shall + have’ before the past participle from the verb. these sentences can be changed into the passive if the active verb has an object


i hope this helps :D thanks
5 0
3 years ago
What is one of the most common ways workers get hurt around machines
-BARSIC- [3]

Answer:

if their body parts stuck in a machine,if machine expl

Explanation:

ode.

4 0
3 years ago
Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th
Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine =1 - \frac{Tl}{Th} \\

where Th =temperature at which the heat enters the engine

Tl is the  temperature of the environment

since both engines have the same thermal capacities <em>n_{th} </em> therefore n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\

We have now that

\frac{-1400}{T}+\frac{T}{300}=0\\

multiplying through by T

-1400 + \frac{T^{2} }{300}=0\\

multiplying through by 300

-420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000}  \\T=648.07k

The temperature T= 648.07k

5 0
3 years ago
Viscous effects are negligible outside of the hydrodynamic boundary layer. (3 points) a. True b. False
Valentin [98]

Answer:

I would say false but I am not for sure

8 0
2 years ago
Which view of an architectural design displays the complex interior features of an object concealed by hidden lines
Klio2033 [76]

Answer:

Cutting-plane View

Explanation:

5 0
3 years ago
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