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Chemistry

1 answer:

Pie3 years ago

4 0

Answer:

A FUSE is a type of conductor which protects the circuit by shorting it down when there is excess flow of current passing through it.

Explanation:

A fuse wire is made up of conducting materials such as alloy of tin and lead that has high resistivity. It has a low melting point of 200°C. It works based on the principle of heating effect of electric current. The functions of fuses include the following:

--> Fuses are made up of thin wire CONDUCTORS which interrupts or breaks the current flow of a circuit when in excess, thereby protecting the circuit from damage.

--> it prevents overload of current. In the event where too many appliances are connected to a single circuit, this can lead to overload which triggers a fuse to terminate the circuit connection.

--> It prevents total black-out: SWITCH-LIKE devices known as CIRCUIT BREAKERS share this function with the fuses. The nearest circuit breaks if any dysfunction occurs in the components of the circuit thereby preventing blackout.

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In the lab, ammonia was mixed with water to form ammonium hydroxide. What is/are the reactant(s)?

Nataly_w [17]

The reactants are what is put into the reaction equation to get the product. In this case water and ammonia are put into the equation to get the product ammonium hydroxide.

The first choice is the correct answer:

Water and ammonia

Hope this helped!

~Just a girl in love with Shawn Mendes

7 0

3 years ago

Read 2 more answers

Calculate the amount of heat released in the combustion of 10.5 grams of Al with 3 grams of O2 to form Al2O3(s) at 25°C and 1 at

ElenaW [278]

Answer : The amount of heat released in the combustion is, 209.5 kJ

Explanation :

First we have to calculate the moles of Al and $O_2$ .

$\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{10.5g}{27g/mole}=0.389moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{3g}{32g/mole}=0.188moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$4Al+3O_2\rightarrow 2Al_2O_3$

From the balanced reaction we conclude that

As, 3 mole of $O_2$ react with 4 mole of $Al$

So, 0.188 moles of $O_2$ react with $\frac{4}{3}\times 0.188=0.251$ moles of $Al$

From this we conclude that, $Al$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Al_2O_3$