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REY [17]
2 years ago
8

How many molecules are present in a drop of ethanol, c2h5oh, of mass 2.3*10^-2.3g? (NA=6.0*10^23mol^-1)

Chemistry
1 answer:
sdas [7]2 years ago
5 0

Answer:

3.0 × 10²⁰ molecules

Explanation:

Given data:

Mass of ethanol = 2.3 × 10⁻²°³ g

Number of molecules = ?

Solution:

Number of moles of ethanol:

Number of moles = mass/ molar mass

Number of moles = 2.3 × 10⁻²°³ g / 46.07 g/mol

Number of moles = 0.05 × 10⁻²°³ mol

Number of molecules:

One mole = 6.022 × 10²³ molecules

0.05 × 10⁻²°³ mol  ×  6.022 × 10²³ molecules / 1 mol

0.30 × 10²⁰°⁷ molecules

3.0 × 10¹⁹°⁷ molecules which is almost equal to 3.0 × 10²⁰ molecules.

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Answer:

B=5522.33kg/h

C=478.11kg/h

Explanation:

Hi! It's a mass balance. First we have to determine the inflow.

mass flow rate = 36000kg / 6h = 6000kg / h

We define the input variable

- input flow (A) = 6000kg / h

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We define output variables.

- skim milk (B)

-creme (C)

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-XgC = percentage of fat in C = 0.45

Then we can start with the balance.

As a general rule, the mass balance is:

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Balance sheet

1) A = B + C

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2) A * XgA = B * XgB + C * XgC

Now we can solve.

We replace and clear B in equation 2

6000kg / h * 0.04 = B * 0.0045 + C * 0.45

B = (240kg / h) /0.045-C*0.45/0.0045

3) B = 53333.33kg / h-C * 100

We replace equation 3 in 1 and clear C

A = B + C

6000kg / h = 53333.33kg / h-C * 100 + C

C=(6000kg/h-53333.33kg/h)/(-99)

C=478.11kg/h

We replace C in equation 3 and calculate B

B = 53333.33kg / h-478.11kg/h * 100

B=5522.33kg/h

Then we have the values ​​of the outflows.

C=478.11kg/h

B=5522.33kg/h

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