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Mandarinka [93]
3 years ago
15

The mass of an egg is approximately 57 g. What is the mass of 3 dozen eggs?

Chemistry
2 answers:
kotegsom [21]3 years ago
6 0
171 because 57x3=171.

BOOM!! 

Shakalaka.






liubo4ka [24]3 years ago
4 0
1 egg = 57 g
1 dozen = 12 eggs
3 dozen = 36 eggs
57 • 36 = mass of 3 Dozen eggs
Mass of 3 Dozen Eggs = 2,052
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Answer:

See explanation

Explanation:

The IUPAC name of aspirin is 2-Acetoxybenzoic acid. It is composed of an acetoxy moiety and a benzoic acid moiety.

The compound can be hydrolysed under prolonged storage conditions to yield acetic acid which causes the vinegar like odour.

Also, one of the products of this hydrolysis bears a phenol group which reacts with FeCl3 to give a purple color.

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A chemist must dilute 97.1 ml of aqueous magnesium fluoride solution until the concentration falls to 389 microMolarity . He'll
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Answer:

0.302L

Explanation:

<em>...97.1mL of 1.21m M aqueous magnesium fluoride solution</em>

<em />

In this problem the chemist is disolving a solution from 1.21mM = 1.21x10⁻³M, to 389μM = 389x10⁻⁶M. That means the solution must be diluted:

1.21x10⁻³M / 389x10⁻⁶M = 3.11 times

As the initial volume of the original concentration is 97.1mL, the final volume must be:

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0.302L

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3 years ago
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6 0
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What is the mass number of sodium
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Answer:

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3 0
3 years ago
Read 2 more answers
2) A common "rule of thumb" -- for many reactions around room temperature is that the
babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$  and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}} $

$\Rightarrow \frac{K_1}{K_2}=  e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

           R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}=  3.479 \times 10^{-6}$

$\Rightarrow  \frac{K_2}{K_1}=  e^{3.479 \times 10^{-6}}$

$\Rightarrow  \frac{K_2}{K_1}=  1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0
3 years ago
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