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mina [271]
3 years ago
15

What mass of aluminum will be deposited on the cathode if an electric current of 0.15 a is run for 4 5s when a solution of al(no

3 )3 is subjected to electrolysis?
Chemistry
1 answer:
marishachu [46]3 years ago
3 0

Mass of Al, m= ?

Molar mass of copper Al, M = 27.0 g/mol

time, t = 45 s

current , c = 0.15 A

valency of Al in Al(NO₃)₃ , z = +3

From Faraday's 1st law,

m = Mct/zF

where F = Faraday constant = 96500 C/mol

Substitute all the values,

m = [(27.0 g/mol) (0.15 A) (45 s)]/ [3] [96500 C/mol]

= 6.3 x 10⁻⁴ g

Mass of aluminium = 6.3 x 10⁻⁴ g

Therefore, mass of aluminium plated on electrode =  6.3 x 10⁻⁴ g

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Answer:

The process of making S'more by adding chocolate bar, gram-crackers, and marshmallows in layers is not a chemical reaction

Explanation:

In a chemical reaction, the substances involved in the reaction are known as the reactants and the substances produced have different physical and chemical properties than those of the reactants and they are known as the products.

The bonds that hold the atoms of the reactants are broken down and rearranged, creating entirely new substances as products. Therefore, energy must be added and/or evolved in any chemical reaction and all reactant atoms should be involved in the reaction.

The change in energy can be sensed as heat change such as increase or decrease in the temperature of the products

Since S'more does not involve any of the above changes that occur in a chemical reaction when the chocolate bar, gram-crackers, and marshmallows are put together, it is not a chemical change or a chemical reaction.

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What mass of HBO2 is produced from the<br> combustion of 139.5 g of B2H6?<br> Answer in units of g.
aliina [53]

Answer:

m_{HBO_2}=441.8gHBO_2

Explanation:

Hello there!

In this case, since the combustion of B2H6 is:

B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O

Thus, since there is 1:2 mole ratio between the reactant and product, the produced grams of the latter is:

m_{HBO_2}=139.5gB_2H_6*\frac{1molB_2H_6}{27.67gB_2H_6} *\frac{2molHBO_2}{1molB_2H_6} *\frac{43.82gHBO_2}{1molHBO_2}

m_{HBO_2}=441.8gHBO_2

Best regards!

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