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3 years ago

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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Who plays a role in the financial activities of a company?

KatRina [158]

Hey,

Who plays a role in the financial activities of a company?

<em>O D. Everyone at the company, including managers and employees</em>

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3 years ago

Technician A says that a defective crankshaft position sensor can cause a no spark condition technician B says that a faulty ign

Aliun [14]

Both tech distributor ignition

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3 years ago

A composite wall is to be used to insulate a freezer chamber at -350C. Two insulating materials are to be used with conductiviti

choli [55]

Answer:

thickness1=1.4m

thickness2=2.2m

convection coefficient=0.33W/m^2K

Explanation:

you must use this equation to calculate the thickness:

L=K(T2-T1)/Q

L=thickness

T=temperature

Q=heat

L1=0.04*(0--350)/10=1.4m

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Then use this equation to calculate the convective coefficient

H=Q/(T2-T1)

H=10/(250-220)=0.33W/m^2K

7 0

3 years ago

The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature

Anit [1.1K]

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

$P_{1}$ = 5 Mpa

$T_{1}$ = 1623°C

= 1896 K

$V_{1}$ = 0.05 $m^{3}$

Also given $\frac{V_{2}}{V_{1}} = 20$

Therefore, $V_{2}$ = 1 $m^{3}$

R = 0.27 kJ / kg-K

$C_{V}$ = 0.8 kJ / kg-K

Also given : $P_{1}V_{1}^{1.25}=C$

Therefore, $P_{1}V_{1}^{1.25}$ = $P_{2}V_{2}^{1.25}$

$5\times 0.05^{1.25}=P_{2}\times 1^{1.25}$

$P_{2}$ = 0.1182 MPa

a). Work transfer, δW = $\frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}$

$\left [\frac{5\times 0.05-0.1182\times 1}{1.25-1} \right ]\times 10^{6}$

= 527200 J

= 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

= $\frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}$

= $\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W$

= $\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200$

= 197.7 kJ

6 0

3 years ago

A water pump delivers 3 hp of shaft power when operating. If the pressure differential between the outlet and the inlet of the p

Natali [406]

Answer:

Mechanical Efficiency = 83.51%

Explanation:

Given Data:

Pressure difference = ΔP=1.2 Psi

Flow rate = $V=8ft^3/s\\$

Power of Pump = 3 hp

Required:

Mechanical Efficiency

Solution:

We will first bring the change the units of given data into SI units.

$P=1.2*6.895 = 8.274KPa\\V=8*0.00283=0.226 m^3/s\\P=3*0.746=2.238KW$

Now we will find the change in energy.

Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.

Thus change in energy is

$=(Mass * change in P)/density\\= \frac{M*P}{p}\\\\$

As we know that Mass = Volume x density

substituting the value

Energy = Volume * density x ΔP / density

Change in energy = Volumetric flow x ΔP

Change in energy = 0.226 x 8.274 = 1.869 KW

Now mechanical efficiency = change in energy / work done by shaft

Efficiency = 1.869 / 2.238

Efficiency = 0.8351 = 83.51%

5 0

3 years ago