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Ugo [173]
1 year ago
6

How many molecules are in 0.25 grams of dinitrogen pentoxide?

Chemistry
2 answers:
ZanzabumX [31]1 year ago
7 0

Answer: 0.014\times 10^{23}molecules

Explanation: Moles can be calculated by using the formula:

\text{no of moles}=\frac{\text{given mass}}{\text{Molecular mass}}

\text{no of moles}=\frac{0.25g}{108g/mol}=0.002moles

1 mole of N_2O_5 contains =6.023\times 10^{23}molecules

0.002 mole of N_2O_5 contains =\frac{6.023\times 10^{23}}{1}\times 0.002=0.014\times 10^{23}molecules

Vesna [10]1 year ago
4 0
In order to get the answer, you have to know the formula of how to convert grams to mole and by that you can get the number of molecules.
The answer is 2 moles of N2O5. This is shown by the solution:
(0.25 g N2O5) (1 mol/ 108 g)=2.31 molecules
The answer is 2 molecules.
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This stadium can hold 100,000, or 1 x 105, people. The number of atoms in a grain of iron is about 1 x 1018. Would you need 1 x 
nata0808 [166]
<h3>Answer:</h3>

1 x 10^13 stadiums

<h3>Explanation:</h3>

We are given that;

1 stadium holds = 1 × 10^5 people

Number of iron atoms is 1 × 10^18 atoms

Assuming the stadium would carry an equivalent number of atoms as people.

Then, 1 stadium will carry 1 × 10^5 atoms

Therefore,

To calculate the number of stadiums that can hold 1 × 10^18 atoms we divide the total number of atoms by the number of atoms per stadium.

Number of stadiums = Total number of atoms ÷ Number of atoms per stadium

                                  = 1 × 10^18 atoms ÷ 1 × 10^5 atoms/stadium

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7 0
1 year ago
A pharmacist–herbalist mixed 100 g lots o St. John’s wort containing the ollowing percentages o the active component hypericin:
Anna [14]

Answer:

strength of hypericin in mixture = 0.42 %

Explanation:

given data

each lot = 100 g

active component hypericin = 0.3%, 0.7%, and 0.25%

solution

we get here percent strength o hypericin in the mixture that is

Hypericin contribution lot 1 =  \frac{0.3}{100} × 100

Hypericin contribution lot 1 =  0.3 g

and

Hypericin contribution lot 2 = \frac{0.3}{100} × 100

Hypericin contribution lot 2 = 0.7 g

and

Hypericin contribution lot 3 = \frac{0.25}{100} × 100  

Hypericin contribution lot 3  = 0.25 g

so

total 300 g mixture of hypericin contain = 0.3 g + 0.7 g + 0.25 g

total 300 g mixture of hypericin = 1.25 g  

so here percent strength o hypericin in mixture is

strength of hypericin in mixture = \frac{1.25}{300} × 100  

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