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Shkiper50 [21]
3 years ago
6

Manu buys two types of wires from a shop - one to put inside a heater as the heating element and the other to use as a fuse wire

for his table lamp. But, he mistakenly puts the fuse wire as the heating coil in a heater.
If the heater draws the same amount or higher amount of current when he switches on the heater, what will probably happen?



A The heater may overheat the room.

B The coil inside the heater may melt and break.

C The heater will start heating the room, but at a very slow pace.

D The coil inside the heater will become an electromagnet instead of heating.


with explanation
Chemistry
1 answer:
joja [24]3 years ago
7 0

There is a major difference in between heating wire and fuse wire

Heating wire has a very high melting point so it can be heated to very high temperature and due to its nature it withstand the temperature and does not melt. They are made up of high melting substance like nichrome.

Fuse has a very low melting point so it cannot be heated to very high temperature and due to its nature it cannot withstand the temperature and get melted easily. When used it melts when the current exceeds a certain limit. That is why it is used a protective aid. Made up of Magnesium etc.


So the correct answer is

the coil inside the heater may melt and break.

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How many grams of silver nitrate will be needed to produce 8.6 g of silver?
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Answer:

13.5g of AgNO3 will be needed

Explanation:

Silver nitrate, AgNO3 contains 1 mole of silver, Ag, per mole of nitrate. To solve this problem we need to convert the mass of Ag to moles. Thee moles = Moles of AgNO3 we need. With the molar mass of AgNO3 we can find the needed mass:

<em>Moles Ag-Molar mass: 107.8682g/mol-</em>

8.6g * (1mol / 107.8682g) = 0.0797 moles Ag = Moles AgNO3

<em>Mass AgNO3 -Molar mass: 169.87g/mol-</em>

0.0797 moles Ag * (169.87g/mol) =

<h3>13.5g of AgNO3 will be needed</h3>

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CHEM HELP!
sweet [91]

So let's convert this amount of mL to grams:

\frac{13.6g}{1mL}*1.2mL=16.32g

Then we need to convert to moles using the molar weight found on the periodic table for mercury (Hg):

\frac{1mole}{200.59g}*16.32g=8.135*10^{-2}mol

Then we need to convert moles to atoms using Avogadro's number:

\frac{6.022*10^{23}atoms}{1mole} *[8.135*10^{-2}mol]=4.90*10^{22}atoms

So now we know that in 1.2 mL of liquid mercury, there are 4.90*10^{22}atoms present.

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