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umka21 [38]
3 years ago
5

What coefficients should be added in order to make this a balanced equation?

Chemistry
1 answer:
Vanyuwa [196]3 years ago
6 0

Answer:

Hello

Coefficients are 1,3,2,1

Explanation

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A compound contains sulfur, oxygen, and chlorine. Analysis shows that it contains by mass 26.95% sulfur and 59.61% chlorine. Wha
sladkih [1.3K]

Answer:

SCl₂

Explanation:

In order to know the empirical formula, we have to follow a series of steps.

Step 1: Divide each percentage by the atomic mass

S: 26.95/32.07 = 0.8403

Cl: 59.61/35.45 = 1.682

Step 2: Divide all the numbers by the smallest one.

S: 0.8403/0.8403 = 1

Cl: 1.682/0.8403 ≈ 2

The empirical formula of the compound is SCl₂.

7 0
3 years ago
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Using the Periodic Table, Which of the following has the largest atomic radius/size?
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3 years ago
Treatment of butanedioic (succinic) anhydride with ammonia at elevated temperature leads to a compound of molecular formula C4H5
artcher [175]

Answer:

The product is Methyl cyanoacetate

Explanation: see structure attached

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3 years ago
If an object has a mass of 20 kg, what is the force of gravity acting on it on earth? A. 196 N B. 32.67 N End of exam C. 1.96 kg
garri49 [273]
Force = mass x gravity

Force = 20 kg x 9.8 m/s²

Force = 196 Newtons

Answer A

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7 0
3 years ago
Number of grams of hydrogen than can be prepared from 6.80g of aluminum​
Anastaziya [24]

Answer:

0.7561 g.

Explanation:

  • The hydrogen than can be prepared from Al according to the balanced equation:

<em>2Al + 6HCl → 2AlCl₃ + 3H₂,</em>

It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.

  • Firstly, we need to calculate the no. of moles of (6.8 g) of Al:

no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.

<em>Using cross multiplication:</em>

2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.

0.252 mol of Al need to react → ??? mol of H₂.

∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.

  • Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:

mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.

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3 years ago
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