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7nadin3 [17]
2 years ago
10

For a school event 1/6 of the athletic field is reversed for the fifth -grade classes the reserved part of the field is divided

equally Among the 4 fifth grade classes in the school what fraction of the whole athletic field is reserved for each fifth class
Chemistry
1 answer:
SVETLANKA909090 [29]2 years ago
3 0

Answer:

\frac{1}{24}

Explanation:

Given:

For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.

To find: fraction of the whole athletic field reserved for each fifth class

Solution:

Fraction of the whole athletic field reserved for four fifth classes = \frac{1}{6}

So, fraction of the whole athletic field reserved for each fifth class = \frac{1}{4}(\frac{1}{6})=\frac{1}{24}

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The frequency of wave is 0.125 Hz.

Explanation:

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"It is an event repeat itself in a given period of time"

The unit of frequency is the Hz . If time is measured in seconds then frequency will be in Hz. Hz is equal to the per second.

Formula:

f = 1/ T

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f = 0.125 Hz

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Potassium iodide reacts with lead(II) nitrate in this precipitation reaction: 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) Wha
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a. 174 mL

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2 KI(aq) + Pb(NO₃)₂(aq) → 2 KNO₃(aq) + PbI₂(s)

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3 years ago
Determine the mass of CuSO4 • 5H20 that must be used to prepare 250mL of 2.01 M CuSO4(aq).
mario62 [17]

Given parameters:

Volume of CuSO₄ = 250mL

Concentration of CuSO₄ = 2.01M

Unknown:

Mass of CuSO₄.5H₂O = ?

To solve this problem, we must write the chemical relationship between both species.;

             CuSO₄.5H₂O  →   CuSO₄ + 5H₂O

Now that we know the expression, it is possible to solve for the unknown mass.

First find the number of moles of CuSO₄;

         Number of moles  = Concentration x Volume

Take 250mL to L so as to ensure uniformity of units;

           Volume  = 250 x 10⁻³L

  Input the parameters and solve for number of moles;

        Number of moles  = 250 x 10⁻³  x  2.01 = 0.5mol

From the equation;

             1 mole of CuSO₄ is produced from 1 mole of CuSO₄.5H₂O  

So  0.5 moles of CuSO₄ will be produced from 0.5 moles of CuSO₄.5H₂O

Now let us find the molar mass of CuSO₄.5H₂O = 63.6 + 32 + 4(16) + 5(2x1 + 16)  = 249.6g/mole

Mass of CuSO₄.5H₂O = number of moles x molar mass

                                      = 0.5 x 249.6

                                     = 124.8g

The mass of CuSO₄.5H₂O is 124.8g

5 0
3 years ago
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