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katovenus [111]
3 years ago
15

A public School district furnishes pencils to its elementary school. The pencils the secretary orders from the district warehous

e each year are packaged in boxes that contain one gross (12 dozen). The average use of pencils is 9.3 pencils per student. If the school's enrollment is 812 students, what is the minimum number of boxes of pencils that should be ordered?
Chemistry
1 answer:
const2013 [10]3 years ago
3 0

Answer:

The minimum number of boxes of pencils to be ordered is 630 boxes.

Explanation:

Since a pupil uses averagely 9.3 pencils

and a box contains 12 pencils,

the school enrollment is also 812

school's enrollment x average use of pencil per student

__________________________________________

            number of pencils in a box

812 x 9.3 = 7551.6

7551.6 /12 = 629.3

Having a total number of 630 boxes of pencils to be ordered.

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2 years ago
Read 2 more answers
A mixture containing nitrogen and hydrogen weighs 3.49 g and occupies a volume of 7.45 L at 305 K and 1.03 atm. Calculate the ma
-BARSIC- [3]

Answer:

Mass percent N₂ = 89%

Mass percent H₂ = 11%

Explanation:

First we <u>use PV=nRT to calculate n</u>, which is the total number of moles of nitrogen and hydrogen:

  • 1.03 atm * 7.45 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305 K
  • n = 0.307 mol

So now we know that

  • MolH₂ + MolN₂ = 0.307 mol

and

  • MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49 g

So we have a <u>system of two equations and two unknowns</u>. We use algebra to solve it:

Express MolH₂ in terms of MolN₂:

  • MolH₂ + MolN₂ = 0.307 mol
  • MolH₂ = 0.307 - MolN₂

Replace that value in the second equation:

  • MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49
  • (0.307-MolN₂) * 2 + MolN₂ * 28 = 3.49
  • 0.614 - 2MolN₂ + 28molN₂ = 3.49
  • 0.614 + 26MolN₂ = 3.49
  • MolN₂ = 0.111 mol

Now we calculate MolH₂:

  • MolH₂ + MolN₂ = 0.307 mol
  • MolH₂ + 0.111 = 0.307
  • MolH₂ = 0.196 mol

Finally, we convert each of those mol numbers to mass, to <u>calculate the mass percent</u>:

  • N₂ ⇒ 0.111 mol * 28 g/mol = 3.108 g N₂
  • H₂ ⇒ 0.196 mol * 2 g/mol = 0.392 g H₂

Mass % N₂ = 3.108/3.49 * 100% = 89.05% ≅ 89%

Mass % H₂ = 0.392/3.49 * 100% = 11.15% ≅ 11%

5 0
3 years ago
If you are using 3.00% (mass/mass) hydrogen peroxide solution and you determine that the mass of solution required to reach the
Maurinko [17]

Answer:

0.004522 moles of hydrogen peroxide molecules are present.

Explanation:

Mass by mass percentage of hydrogen peroxide solution = w/w% = 3%

Mass of the solution , m= 5.125 g

Mass of the hydrogen peroxide = x

w/w\% = \frac{x}{m}\times 100

3\%=\frac{x}{5.125 g}\times 100

x=\frac{3\times 5.125 g}{100}=0.15375 g

Mass of hydregn pervade in the solution = 0.15375 g

Moles of hydregn pervade in the solution :

=\fraC{ 0.15375 g}{34 g/mol}=0.004522 mol

0.004522 moles of hydrogen peroxide molecules are present.

5 0
3 years ago
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