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Yakvenalex [24]
3 years ago
11

What do these red/yellow/orange-precipitate reactions have in common? (choose

Chemistry
1 answer:
aev [14]3 years ago
8 0

Answer:

3 is the correct I think so

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How many atoms are present in 3 moles of chromium?
Murljashka [212]

Answer:

1.80 x 10^24 atoms

Explanation:

3moles × 6.022×10^23 atoms/mole

3 0
3 years ago
A chemist adds of a zinc nitrate solution to a reaction flask. Calculate the mass in kilograms of zinc nitrate the chemist has a
Ray Of Light [21]

Answer:

5.3 × 10⁻³ kg

Explanation:

There is some info missing. I think this is the original question.

<em>A chemist adds 135.0 mL of a 0.21 M zinc nitrate (Zn(NO₃)₂) solution to a reaction flask. Calculate the mass in kilograms of zinc nitrate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.</em>

<em />

We have 135.0 mL of a 0.21 M zinc nitrate (Zn(NO₃)₂) solution. The moles of zinc nitrate are:

0.1350 L × 0.21 mol/L = 2.8 × 10⁻² mol

The molar mass of zinc nitrate is 189.36 g/mol. The mass corresponding to 2.8 × 10⁻² moles is:

2.8 × 10⁻² mol × 189.36 g/mol = 5.3 g

1 kilogram is equal to 1000 grams. Then,

5.3 g × (1 kg/1000 g) = 5.3 × 10⁻³ kg

8 0
3 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
What is the sum of 16+ -20?
mr Goodwill [35]

Answer:

-4

Explanation:

PLz give me brainliest I worked hard thank u

6 0
3 years ago
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How many times greater is the rate of effusion of oxygen gas than that of carbon dioxide gas at the same temperature and pressur
AlexFokin [52]
C.) 1.173. Hope this helps you.

6 0
3 years ago
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