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3 years ago

NEED HELP ASAP! Write a statement explaining your claim about whether or not the equation is correctly balanced.

Chemistry

2 answers:

cupoosta [38]3 years ago

8 0

Answer:

I think that it it correctly balanced that is my opinion and, because the way it is set up, that the answer will tell you weather or not it is correctly balanced or not.

Kisachek [45]3 years ago

4 0

Evaluating the number of atoms of each element on the product and reactant side, we can conclude that the student's equation isn't balanced.

Given the chemical reaction :

$Al_{4}C_{3} + 6H_{2}O ---> 3CH_{4} + 4Al(OH)_{3}$

For a balanced equation, the number of atoms of each element in the product side must be equal to that on the reactant side.

Number of hydrogen :

Reactant side = (6 × 2) = 12

Product side = (6 × 2) + (6 × 2) = 12 + 24 = 24

Product side ≠ reactant side ; Hence, the equation isn't balanced

The balanced equation for the reaction goes thus :

$Al_{4}C_{3} + 12H_{2}O ---> 3CH_{4} + 4Al(OH)_{3}$

Reactant side :

Al = 4

H = (12 × 2) = 24

O = 12

C = 3

Product side :

H = (6 × 2) + (6 × 2) = 12 + 24 = 24

Al = 4

O = 4 × 3 = 12

C = 3

Therefore, the student's equation isn't balanced, the balanced version of the equation has been given .

Learn more :brainly.com/question/2720947

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4 years ago

f 4.2 moles of copper metal react with 6.3 moles of silver nitrate, how many moles of silver metal can be formed, and how many m

alex41 [277]

1. Balance the reaction
Cu + 2AgNO3 → Cu(NO3)2 + 2Ag
2. Find the limiting [divide moles of reactant by their balancing numbers]
Cu: 4.2/1 =4.2
AgNO3: 6.3/2= 3.15
AGNO3 is the limiting
How many moles of silver metal will be formed?
The ratio between the limiting reactant and silver is 2:2 or 1:1
therefore 6.3 moles of silver will be formed.

How many moles of excess reactant will be left?
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3 years ago

The covalent catalytic mechanism of an enzyme depends on a single active site Cys whose pK is 8. A mutation in a nearby residue

Nimfa-mama [501]

Answer: The reaction rate will INCREASE

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3 years ago

A compound containing xenon and fluorine was prepared by shining sunlight on a mixture of Xe (0.526 g) and excess F2 gas. If you

Galina-37 [17]

Answer:

$XeF_2$

Explanation:

Hello!

In this case, according the law of conservation of mass, we can see that, since the total mass of the Xe-F compound is 0.678 g and that of xenon is 0.526 g, we have 0.152 g as the mass of reactants equal the mass of the products. It means that we can compute the moles of each atom in the compound as shown below:

$n_{Xe}=0.526g/131.3g/mol=0.004mol\\\\n_F=0.152g/19.0g/mol=0.008mol$

Now we divide the moles of both reactants by the moles of xenon as those are the fewest ones in order to find their subscripts in the empirical formula:

$Xe=0.004/0.004=1\\\\F=0.008/0.004=2$

Thus, the empirical formula is:

$XeF_2$

Best regards!

4 0

3 years ago