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ratelena [41]
3 years ago
11

Describe how the energy consumption of appliances is dependent on the power rating of an appliance and the length of time that t

he appliance is used for
Chemistry
1 answer:
lawyer [7]3 years ago
5 0

Answer:

  • Different appliances and machines are analyzed on the basis of the energy that each of the machine or appliance consumes in a given time. As, we know that the voltage is the amount or level of difference between the energy possessed by the different points or devices.
  • As, the required amount of energy required to run the machine or any device is consumed from the reservoir of the potential reservoir, as there is a certain scale of time, T and power,P on which the different machines are analyzed in more proper way and find its total efficiency in the mean time,T.
  • However, the power rating or power consumption is the amount of voltage dissipated during the working of the device and the amount of current, I flowing through the device's system. As, if we analyze the over all working of the system on the given scale of time,T. As the level of energy consumed and the current passed through these system give us the power ratings,P(rating) on a given scale of time,T.
  • The unit will be watt per hour for the power in the given amount of time.

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After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2 is
Tanzania [10]

Answer:

Ksp = 8.8x10⁻⁵

Explanation:

<em>Full question is:</em>

<em>After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2+ is 2.8 × 10–2 M. What is Ksp for PbCl2?</em>

<em />

When an excess of PbCl₂ is added to water, Pb²⁺ and Cl⁻ ions are produced following Ksp equilibrium:

PbCl₂(s) ⇄ Pb²⁺ + 2Cl⁻

Ksp = [Pb²⁺] [Cl⁻]²

If an excess of PbCl₂ was added, an amount of Pb²⁺ is produced (X) and twice Pb²⁺ is produced as Cl⁻ (2X):

Ksp = [X] [2X]²

Ksp = 4X³

As X is the amount of Pb²⁺ = 2.8x10⁻²M:

Ksp = 4(2.8x10⁻²)³

<h3>Ksp = 8.8x10⁻⁵</h3>
4 0
4 years ago
If 4.50 g of (nh4)2so4 is dissolved in enough water to form 250. ml of solution, what is the molarity of the solution?
natima [27]
The molarity of a solution is the number of moles of a substance in one liter of that substance. 
The molar mass of ammonium sulfate (NH4)2SO4 is 132.14 grams/mole
Calculate the moles of ammonium sulfate:
(4.50 grams)/(132.14 grams/mole) = 0.0341 moles of ammonium sulfate
convert mL to Liters 250. mL becomes 0.250 liters
Take the number of moles over the number of liters
0.0341 moles / 0.250 liters = 0.136 molar or 0.136M = molarity of the solution
8 0
3 years ago
What is white light?
muminat
White light is what the eye sees when wavelengths of all colours reach the eye.

It is a combination of Red, Blue and Green wavelengths of light, that is perceived as white.
6 0
4 years ago
Read 2 more answers
What are the forces that hold sodium and chloride ions together?
Phantasy [73]
Sodium is a metal, Chloride is a non-metal.

Right off the bat, you know that in order for both of these atoms to achieve a full valence shell that the metal has to lose electrons, and the non-metal has to gain them.

Therefore, you have the transfer of electrons in this bond in order to form ions.
Na+ and Cl-. This transfer of electrons in a bond is called an {{ Ionic Bond}}
6 0
3 years ago
During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce
swat32

Answer: 16.32 g of O_2 as excess reagent are left.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol

\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)  

According to stoichiometry :

2 moles of SO_2 require = 1 mole of O_2

Thus 0.34 moles of SO_2 will require=\frac{1}{2}\times 0.34=0.17moles  of O_2

Thus SO_2 is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

Moles of O_2 left = (0.68-0.17) mol = 0.51 mol

Mass of O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g

Thus 16.32 g of O_2 as excess reagent are left.

3 0
3 years ago
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