Calculate the weight of one medium egg if a dozen medium eggs has an average weight of 21 ounces

Chemistry

2 answers:

Scrat [10]3 years ago

7 0

Answer:

1.75 ounces

Explanation:

Divide 21 ounces by the number of eggs:

21/12 = 1.75

mash [69]3 years ago

4 0

<u>Answer:</u> The weight of one medium egg is 1.75 ounce.

<u>Explanation:</u>

We are given:

Weight of dozen medium eggs = 21 ounces

We know that:

1 dozen = 12 eggs

Applying unitary method:

12 dozen medium eggs has a weight of 21 ounces

So, 1 medium egg will have a weight of $\frac{21}{12}\times 1=1.75$ ounce

Hence, the weight of one medium egg is 1.75 ounce.

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What is the bronsted acid of NO2- + H2O HNO2 + OH-

sammy [17]

<h3><u>Answer</u>;</h3>

H2O -Bronsted Acid

<h3><u>Explanation;</u></h3>

Bronsted-Lowry acids are H+ donors , while Bronsted-Lowry bases are H+ acceptors .
A reaction of a Bronsted-Lowry acid and a Bronsted base is a neutralization reaction that is characterized by H+ transfer.

The above reaction is an example of base ionization or dissociation where;

B (aq) + H2O (l) → BH+ (aq) + OH– (aq)

That is; Base + Acid will give a conjugate acid + hydroxide ion

In our case; NO2- + H2O → HNO2 + OH- ; H2O is the H+ donor and thus, it is a Bronsted Acid.

8 0

3 years ago

Pentaborane-9, B5H9, is a colorless, highly reactive liquid that will burst into flame when exposed to oxygen. The reaction is 2

mina [271]

<u>Answer:</u> The amount of energy released per gram of $B_5H_9$ is -71.92 kJ

<u>Explanation:</u>

For the given chemical reaction:

$2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]$

Taking the standard enthalpy of formation:

$\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol$