Hydrogen gas can be produced by reacting a mixture of methane and steam in the

What is the average atomic mass of chlorine if the mass of one isotope is 34.97 amu and has a p?er cent abundance of 75.77%. The

pishuonlain [190]

Answer : The average atomic mass of chlorine is, 32.37 amu

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Mass of isotope 1 = 34.97 amu

Percentage abundance of isotope 1 = 75.77 %

Fractional abundance of isotope 1 = 0.7577

Mass of isotope 2 = 36.97 amu

Percentage abundance of isotope 2 = 24.23 %

Fractional abundance of isotope 2 = 0.2423

Now put all the given values in above formula, we get:

$\text{Average atomic mass of element}=\sum[(34.97\times 0.7577)+(24.23\times 0.2423)]$

$\text{Average atomic mass of element}=32.37amu$

Therefore, the average atomic mass of chlorine is, 32.37 amu

8 0

3 years ago

Cu20(s) + C(s) - 2Cu(s) + CO(g)

Romashka-Z-Leto [24]

Answer:

That means Cu2O is limiting reagent and C is excess reagent

Explanation:

Based on the reaction, 1 mole of Cu2O reacts per mole of C. The ratio of reaction is 1:1.

To solve this question we need to convert the mass of each reactant to moles. The reactant with the lower amount of moles is limiting reactant and the excess reactant is the reactant with the higher number of moles.

<em>Moles Cu2O -Molar mass: 143.09 g/mol-</em>

114.2g Cu2O * (1mol / 143.09g) = 0.798 moles Cu2O

<em>Moles C -Molar mass: 12.01g/mol-</em>

11.1g C * (1mol / 12.01g) = 0.924 moles C

<h3>That means Cu2O is limiting reagent and C is excess reagent</h3>

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6 0

3 years ago

The line that divides the western and eastern hemispheres is the

RideAnS [48]

Prime meridian (hope this helps)

6 0

4 years ago

Read 2 more answers

Help please At which boundary type would mountains, such as the Himalayans, most likely form?

Liono4ka [1.6K]

Answer:

convergent is the answer

6 0

3 years ago

. In an experiment, 1.90 g of NH3 reacts with 4.96 g of O2. 4NH3(g) + 5O2(g) ⟶ 4NO(g) + 6H2O(g) (i) Which is a limiting reactant

Simora [160]

1. The limiting reactant in the reaction is NH₃

2. The mass of the excess reactant remaining is 0.49 g

3. The mass of NO produced from the reaction is 3.35 g

<h3>Balanced equation </h3>

4NH₃ + 5O₂ —> 4NO + 6H₂O

Molar mass of NH₃ = 14 + (3×1) = 17 g/mol

Mass of NH₃ from the balanced equation = 4 × 17 = 68 g

Molar mass of O₂ = 16 × 2 = 32 g/mol

Mass of O₂ from the balanced = 5 × 32 = 160 g

Molar mass of NO = 14 + 16 = 30 g/mol

Mass of NO from the balanced equation = 4 × 30 = 120 g

SUMMARY

From the balanced equation above,

68 g of NH₃ reacted with 160 g of O₂ to produce 120 g of NO

<h3>1. How to determine the limiting reactant </h3>

From the balanced equation above,

68 g of NH₃ reacted with 160 g of O₂

Therefore,

1.90 g of NH₃ will react with = (1.90 × 160) / 68 = 4.47 g of O₂

From the calculation made above, we can see that only 4.47 g out of 4.96 g of O₂ given is needed to react completely with 1.90 g of NH₃.

Therefore, NH₃ is the limiting reactant.

<h3>2. How to determine the mass of the excess reactant remaining </h3>

Mass of excess reactant (O₂) given = 4.96 g
Mass of excess reactant (O₂) that reacted = 4.47 g

Mass of excess reactant (O₂) remaining =?

Mass of excess reactant (O₂) remaining = 4.96 – 4.47

Mass of excess reactant (O₂) remaining = 0.49 g

<h3>3. How to determine the mass of NO produced </h3>

In this case, the limiting reactant (NH₃) will be used.

From the balanced equation above,

68 g of NH₃ reacted to produce 120 g of NO

Therefore,

1.90 g of NH₃ will react to produce = (1.90 × 120) / 68 = 3.35 g of NO.

Thus, 3.35 g of NO were obtained from the reaction.

Learn more about stoichiometry:

brainly.com/question/14735801

5 0

2 years ago