Даю 90 баллов, помогите

Explain why energy sources do not have 100% efficiency. Why do you think some have lower efficiencies?

bearhunter [10]

Answer:

Energy sources do not have 100% efficiency because the processes of energy conversion to usable forms involves energy losses.

Some have lower efficiencies due to; energy losses in form of heat during conversion, poor technology applied during conversion of energy and lack of desire equipment to use in the energy conversion system.

Explanation:

The desired form of energy for use is derived from conversion of energy from the source using an energy converter into another form which is usable. The efficiency of the energy converter is calculated as;

л = output energy/input energy

The efficiency of energy is limited to the cost of equipment required for conversion from energy source by the energy converter to a form which is usable. Additionally, because energy sources are scarce, the technology to use in energy conversion is a factor affecting energy efficiency in that high efficiency will require advanced technology with better equipment leading higher costs of that energy form. when heat losses are involved during energy conversion, efficiency lowers, thus its better if such losses are used as energy input in another system.

5 0

2 years ago

Sodium sulfate is slowly added to a solution containing 0.0500 M Ca 2 + ( aq ) and 0.0390 M Ag + ( aq ) . What will be the conce

levacccp [35]

The given question is incomplete. The complete question is as follows.

Sodium sulfate is slowly added to a solution containing 0.0500 M $Ca^{2+}(aq)$ and 0.0390 M $Ag^{+}(aq)$ . What will be the concentration of $Ca^{2+}$ (aq) when $Ag_{2}SO_{4}(s)$ begins to precipitate? What percentage of the $Ca^{2+}(aq)$ can be separated from the Ag(aq) by selective precipitation?

Explanation:

The given reaction is as follows.

$Ag_{2}SO_{4} \rightleftharpoons 2Ag^{+} + SO^{2-}_{4}$

$[Ag^{+}]$ = 0.0390 M

When $Ag_{2}SO_{4}$ precipitates then expression for $K_{sp}$ will be as follows.

$K_{sp} = [Ag^{+}]^{2}[SO^{2-}_{4}]$

$1.20 \times 10^{-5} = (0.0390)^{2} \times [SO^{2-}_{4}]$

$[SO^{2-}_{4}]$ = 0.00788 M

Now, equation for dissociation of calcium sulfate is as follows.

$CaSO_{4} \rightleftharpoons Ca^{2+} + SO^{2-}_{4}$

$K_{sp} = [Ca^{2+}][SO^{2-}_{4}]$

$4.93 \times 10^{-5} = [Ca^{2+}] \times 0.00788$

$[Ca^{2+}]$ = 0.00625 M

Now, we will calculate the percentage of $Ca^{2+}$ remaining in the solution as follows.

$\frac{0.00625}{0.05} \times 100$

= 12.5%

And, the percentage of $Ca^{2+}$ that can be separated is as follows.

100 - 12.5

= 87.5%

Thus, we can conclude that 87.5% will be the concentration of $Ca^{2+}(aq)$ when $Ag_{2}SO_{4}(s)$ begins to precipitate.

4 0

2 years ago

Copper is formed when aluminum reacts with copper (II) sulfate in a single-replacement reaction. How many moles of copper can be

serg [7]

Answer:

The answer to your question is the limiting reactant is CuSO₄ and 0.975 moles of Cu were obtained

Explanation:

moles of Copper = ?

mass of Aluminum = 29 g

mass of CuSO₄ = 156 g

Limiting reactant = ?

Balanced Chemical reaction

3 CuSO₄ + 2 Al ⇒ 3 Cu + Al₂(SO₄)₃

Calculate the moles of reactants

CuSO₄ = 64 + 32 + (16 x 4) = 160g

Al = 27 g

160 g of CuSO₄ ----------------- 1 mol

156 g ----------------- x

x = (156 x 1) / 160

x = 0.975 moles

27 g of Al -------------------------- 1 mol

29 g of Al -------------------------- x

x = (29 x 1)/27

x = 1.07 moles

Calculate proportions to find the limiting reactant

Theoretical 3 moles CuSO₄/2 moles Al = 1.5 moles

Experimental 0.975 moles CuSO₄/1.07 moles = 0.91

The experimental proportion was lower than the theoretical proportion that means that the limiting reactant is CuSO₄.

3 moles of CuSO₄ ------------------ 3 moles of Cu

0.975 moles of CuSO₄ --------------- x

x = (0.975 x 3)/3

x = 0.975 moles of Cu were obtained.

3 0

2 years ago

Read 2 more answers

What mass of Cu(s) is electroplated by running 24.5A of current through a Cu2+(aq)solution for 4.00 h?Express your answer to thr

Masja [62]

Answer: 116 g of copper

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 24.5A

t= time in seconds = 4.00 hr = $4.00\times 3600s=14400s$ (1hr=3600s)

$Q=24.5A\times 14400s=352800C$

$Cu^{2+}+2e^-\rightarrow Cu$

$2\times 96500C=193000C$ of electricity deposits 63.5 g of copper.

352800 C of electricity deposits = $\frac{63.5}{193000}\times 352800=116g$ of copper.

Thus 116 g of Cu(s) is electroplated by running 24.5A of current

Thus remaining in solution = (0.1-0.003)=0.097moles

8 0

2 years ago

All the elements of a family in the periodic table have what feature in common?a)they all have similar chemical properties.b)the

Vadim26 [7]

Answer: D. They all have the same number of electrons in the electron cloud

5 0

2 years ago

Read 2 more answers