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faltersainse [42]
2 years ago
8

The Chemistry Cafe was out of bread. The cook went next door to the bakery and bought a loaf of bread which has 33 slices. Then,

she sells 12 sandwiches (two pieces of bread per sandwich). How much bread did she have left? Explain your answer.​
Chemistry
2 answers:
maria [59]2 years ago
4 0

In the question, we are told that there are;

  • A loaf containing 33 slices
  • A loaf containing 33 slices A package of cheese containing 15 slices

We also know that he is making a sandwich that has 2 pieces of both cheese and bread.

Hence;

Total number of bread and cheese = 33 + 15.

Each loaf should have two pieces of each bread and the cheeses make a total of four pieces.

Therefore he can make = 33 + 15/4 = 12 sandwiches.

scoundrel [369]2 years ago
4 0
The cook uses 2 slices of bread per sandwich. Which is equivalent to (the slices of bread) x (the total number of sandwiches) = total number of bread slices used. In this case 2 x 12 =24 slices used. Since the loaf of bread had 33, subtract the slices of bread used, 24. This results in 33-24= 9 slices of bread remaining.
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How many moles of n are in 0.187 g of n2o?
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We know that the molar mass of N is 14 and O is 16, therefore the molar mass of N2O is:

molar mass N2O = 14 * 2 + 16 = 44 g/mol

 

The number of moles:

moles N2O = 0.187 / 44

moles N2O = 0.00425 mol

 

There are 2 moles of N per 1 mole of N2O hence:

moles N = 0.00425mol * 2

<span>moles N = 0.0085 mol</span>

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You are asked to prepare 500 mL 0.200 M acetate buffer at pH 5.00 using only pure acetic acid ( MW=60.05 g/mol, pKa=4.76), 3.00
Vilka [71]

Answer:

You need to weight 6,005 g of acetic acid

Explanation:

Using Henderson-Hasselbalch formula you will obtain:

5,00 = 4,76 +log₁₀ \frac{[Ac^-]}{[Acac]}

<em>Where Ac⁻ is the salt of acetic acid (Acac).</em>

Solving:

1,738 = \frac{[Ac^-]}{[Acac]} <em>(1)</em>

Also, yo know that:

0,200 M = [Ac⁻] + [Acac] <em>(2)</em>

Replacing (2) in (1):

[Acac] = 0,0730 M.

Thus:

[Ac⁻] = 0,127 M

The moles of each compound are:

Acac = 0,0730 M × 0,500 L = <em>0,0365 mol</em>

Ac⁻ = 0,127 M × 0,500 L = <em>0,0635 mol</em>

To prepare these moles it is necessary to use:

Acac + NaOH → AcNa + H₂O

The initial moles of Acac must be:

0,0365 moles + 0,0635 moles = 0,100 moles

<em>To obtain 0,0635 moles of Ac⁻ you need to take this quantity of NaOH moles.</em>

Thus, to obtain a acetate buffer of 5,00 you need to add 0,100 moles of acetic acid and 0,0635 moles of NaOH because This NaOH will react with acetic acid producing 0,0635 moles of Ac⁻ and surplus 0,0365 moles of acetic acid.

Now, to obtain 0,100 moles of acetic acid from pure acetic acid:

0,100 moles × \frac{60,05 g}{1 mol} = <em>6,005 g</em>

<em>You need to weight 6,005 g of acetic acid</em>

I hope it helps!

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