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7nadin3 [17]
3 years ago
15

Compare the mass of the Mg ribbon with the mass of the magnesium oxide. Notice that the the mass of the magnesium oxide is great

er than the mass of the Mg. How do you account for this apparent increase in mass?
Chemistry
1 answer:
Artist 52 [7]3 years ago
4 0

Answer:

Magnesium oxide is a binary compound of magnesium and oxygen while magnesium ribbon consists only of magnesium atoms.

Explanation:

The burning of magnesium in oxygen is a chemical change. It produces magnesium oxide having greater mass than magnesium ribbon. The greater mass results from the fact that the chemical reaction has added another element to the sample- oxygen. The mass of magnesium ribbon is the mass of magnesium atoms alone but in magnesium oxide, we consider the masses of magnesium and oxygen atoms making magnesium oxide heavier than magnesium ribbon.

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Rory answered that absolute zero is 0°K. Is this answer correct? Explain why or why not.
Ierofanga [76]

Answer:

Yes, this answer is correct.

Explanation:

Absolute zero is -273.15 degrees celsius

T= Tc+273.15

T is the kelvin temperature

Tc is the temperature in degrees celsius

substitute;

T= -273.15 + 273.15

T=0 K

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3 years ago
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Answer: The correct answer is in chemical bonds

Explanation:

When coal is burnt, these components burn and release energy. The energy released is by the chemical reaction between the constituents and oxygen

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3 years ago
What is the law of multiple proportions
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7 0
3 years ago
An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni
Svetllana [295]

<u>Answer:</u> The pH of acid solution is 4.58

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}    .....(1)

  • <u>For KOH:</u>

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol

  • <u>For propanoic acid:</u>

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol

The chemical reaction for propanoic acid and KOH follows the equation:

                 C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O

<u>Initial:</u>          0.1372         0.04514  

<u>Final:</u>           0.09206          -                0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L     (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})

pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})

We are given:  

pK_a = negative logarithm of acid dissociation constant of propanoic acid = 4.89

[C_2H_5COOK]=\frac{0.04514}{0.26594}

[C_2H_5COOH]=\frac{0.09206}{0.26594}

pH = ?  

Putting values in above equation, we get:

pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58

Hence, the pH of acid solution is 4.58

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