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3 years ago

What is the maximum mass of b4c that can be formed from 2.00 moles of boron(iii) oxide?

Chemistry

1 answer:

anzhelika [568]3 years ago

3 0

The maximum mass of B₄C that can be formed from 2.00 moles of boron (III) oxide is 55.25 grams.

<h3>What is the stoichiometry?</h3>

Stoichiometry of the reaction gives idea about the relative amount of moles of reactants and products present in the given chemical reaction.

Given chemical reaction is:

2B₂O₃ + 7C → B₄C + 6CO

From the stoichiometry of the reaction, it is clear that:

2 moles of B₂O₃ = produces 1 mole of B₄C

Now mass of B₄C will be calculated by using the below equation:

W = (n)(M), where

n = moles = 1 mole
M = molar mass = 55.25 g/mole

W = (1)(55.25) = 55.25 g

Hence required mass of B₄C is 55.25 grams.

To know more about stoichiometry, visit the below link:

brainly.com/question/25829169

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Explain how engineers and scientists may work together to produce a protein in the lab.

murzikaleks [220]

The scientists would do biological studies of how the protein breakdown and combines with the muscles the engineers with then create a delivery system to get the protein to the muscle quicker and more effectively

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135-g sample of a metal requires 2.50 kJ to change its temperature from 19.5 C to 100.0 C what is the specific heat of this meta

Harman [31]

The temperature increase is from 19.5 to 100 degrees centigrade or 80.5 degrees centigrade. The calorie increase is 2.50 x 1000 x 0.238902957619 or a total of 597.25 calories. 597.25/80.5 = 7.419 calories per degree centigrade. 7.419/135 grams = 0.0549 calories/gram/degree centigrade. The conversion from kilo joules involves multiplying the calories per joule x 1000 to get the number of calories in one kilo joule and then by the 2.5.

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3 years ago

What mass of aluminum is needed to produce 0.500 mole of aluminum chloride?

3241004551 [841]

Answer: " 13.5 g Al " ;

→ that is: "13.5 grams of aluminum."

____________________________

Explanation:

____________________________

Note: What is missing from the question is the "balanced chemical equation" for the "chemical reaction" that contains:

The reactants: "aluminum (Al) " ; and "chlorine (Cl) " ; and:

The product: "aluminum choloride (AlCl₃) " .

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The "balanced chemical equation" is:

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2 Al + 3 Cl₂ → 2 AlCl₃ ;

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Note: The molecular weight of "aluminum (Al)" is: " 26.98 g /mol " .

____________________________

So: We call solve using a technique known as: "dimensional analysis" :

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0.500 mol AlCl₃ * $(\frac{2mol Al}{2mol AlCl_{3} }) * (\frac{26.98g Al}{1 mol Al}) = ?$

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Note: The units of "mol AlCl₃" cancel out to "1' ; and:

The units of "mol Al" cancel out to "1" ; and we are left with:

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" $\frac{(0.500 * 2 * 26.98)}{2}$ g Al ["grams of aluminum"] ;

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Note: We can "cancel out the "2's" ; since "2/2 = 1 " ; and we have:

→ (0.500 * 26.98) g Al ;

= 13.49 g Al ;

→ Round to 3 (Three) significant figures;

→ Since: "0.500" has 3 (Three) significant figures:

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= 13.5 g Al ; that is: "13.5 grams of aluminum."

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Hope this is helpful!

Best wishes to you in your academic pursuits—and within the "Brainly" community!

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3 years ago

Two elements in the same period have the same number of _____ _____ in their electron clouds.

shutvik [7]

Answer:

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Explanation:

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Write the net ionic equation for the reaction that occurs when equal volumes of 0.546 M aqueous acetylsalicylic acid (aspirin) a

Tatiana [17]

Answer:

$C_9H_8O_4+C_2H_3O_2^-\rightarrow C_2H_4O_2+C_9H_7O_4^-$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to figure out the required net ionic equation by firstly writing out the complete molecular equation between aspirin and sodium acetate:

$C_9H_8O_4+NaC_2H_3O_2\rightarrow C_2H_4O_2+NaC_9H_7O_4$

Whereas acetic acid and sodium acetylsalicylate are formed. Now, we write the complete ionic equation whereby sodium acetate and sodium acetylsalicylate are ionized because they are salts yet neither aspirin nor acetic acid are ionized as they are weak acids:

$C_9H_8O_4+Na^++C_2H_3O_2^-\rightarrow C_2H_4O_2+Na^++C_9H_7O_4^-$

Finally, for the net ionic equation we cancel out the sodium spectator ions to obtain:

$C_9H_8O_4+C_2H_3O_2^-\rightarrow C_2H_4O_2+C_9H_7O_4^-$

Regards!

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3 years ago