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Korvikt [17]
3 years ago
14

How many molecules are in 68.0 g of H2S

Chemistry
1 answer:
Kryger [21]3 years ago
3 0

The correct answer is 12.044 × 10²³ molecules.  

The molecular mass of H₂S is 34 gram per mole.  

Number of moles is determined by using the formula,  

Number of moles = mass/molecular mass

Given mass is 68 grams, so no of moles will be,  

68/34 = 2 moles

1 mole comprises 6.022 × 10²³ molecules, therefore, 2 moles will comprise = 6.022 × 10²³ × 2

= 12.044 × 10²³ molecules.  


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Vinil7 [7]

Answer: F. Electrons

Explanation: hope it helped    .u.

3 0
3 years ago
Read 2 more answers
usando el concepto de valencia para los elementos S, P Y Br predice las formulas de los compuestos mas simples formados por esto
hram777 [196]

Explanation:

Teniendo en cuenta los numeros de oxidacion negativos de cada uno

P:fosforo(-3)

S:Azufre (-2)

Br:Bromo (-1)

Y el H: hidrogeno con una valencia

positiva de +1

los compuesto que se formaran son los siguientes

1 H3P= fosfuro de hidrogeno

2 H2S= sulfuro de hidrogeno

3 HBr= Acido bromhidrico

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6 0
2 years ago
A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a
Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

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,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = \frac{x}{100}

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = \frac{10}{100}

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] %  = (90 - x) %

Fractional abundance of middle-weight isotope = \frac{(90-x)}{100}

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

48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]

x=78\%

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0
3 years ago
Read 2 more answers
A sample of a food label is shown below.
White raven [17]

Answer:

A) It's correctly written

B) 77%

C) 835 calories

Explanation:

A) From online sources, we have number of calories as follows;

Fats: 9 calories per gram

Protein; 4 calories per gram

Carbs; 4 calories per gram

Total calories for each;

Total fat = 3 × 9 = 27 calories

Total protein = 3 × 4 = 12 calories

Total carbs = 32 × 4 = 128 calories

(sugar and dietary Fibre are classified as carbohydrates and so total carbs takes care of their calories).

Thus, total number of calories per serving = 27 + 12 + 128 = 167 calories per serving which is same as what is given.

B) percent from carbohydrates per serving = total calories from carbs/total number of calories per serving × 100% = 128/167 × 100% ≈ 77%

C) One box contains 5 servings. Thus total number of calories per box = 167 × 5 = 835 calories

8 0
3 years ago
Which of these describes an endothermic reaction?
neonofarm [45]
B) energy is absorbed by the reaction
is right answer.
3 0
3 years ago
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