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Pepsi [2]
3 years ago
15

Slicing an average loaf of bread gives 13 medium thickness slices. Each slice of bread is to be coated with 0.0500 oz of mayonna

ise. Each sandwich of two bread slices contains one slice of ham and one slice of cheese. Half of the sandwiches need to contain a sliced olive, the other half have no olive. How many loaves of bread and how many ounces of mayonnaise are needed to make all of the sandwiches if you need 231 ham and cheese sandwiches with an olive?
Chemistry
1 answer:
AlexFokin [52]3 years ago
8 0

Answer:

36 loaves of bread and 18.48 ounces of mayonnaise are needed to make all of the sandwiches.

Explanation:

Number of slices of ham = 231

Given that each sandwich has 1 slice of ham.

Then 231 slices of ham will be in :

1\times 231 =231 sandwiches

A single sandwich is made up of 2 bread slices , Then bread slices in 231 sandwiches will be :

231\times 2=462

Each bread slice is applied with 0.0500 ounce of mayonnaise.

Then 462 bread slices will have :

462\times 0.0500 ounces=18.48 Oz

1 bread loaf = 13 bread slices

Then breads loves in 462  bread slices :

\frac{462}{13}=35.54\approx 36

36 loaves of bread and 18.48 ounces of mayonnaise are needed to make all of the sandwiches.

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Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water
madreJ [45]

Explanation:

The given data is as follows.

         Temperature of dry bulb of air = 25^{o}C

          Dew point = 10^{o}C = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

        Partial pressure of water vapor (P_{a}) = vapor pressure of water at a given temperature (P^{s}_{a})

Using Antoine's equation we get the following.

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}

                               = 0.17079

                   P^{s}_{a} = 1.18624 kPa

As total pressure (P_{t}) = atmospheric pressure = 760 mm Hg

                                   = 101..325 kPa

The absolute humidity of inlet air = \frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                  \frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                 = 0.00735 kg H_{2}O/ kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg H_{2}O/ kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

                 0.07 kg - 0.00735 kg

              = 0.06265 kg H_{2}O for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

                0.06265 kg \times 100

                  = 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

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3 years ago
Suppose that X represents an arbitrary cation and that Y represents an anionic species. Using the charges indicated in the super
dmitriy555 [2]

Answer:

See explanation.

Explanation:

Hello!

In this case, when having the cationic and anionic species with the specified charges, in order to abide by the net charge rule, we need to exchange the charges in the form of subscripts and without the sign, just as shown below:X^{m+}Y^{n-}\rightarrow X_nY_m

Thus, for all the given combinations, we obtain:

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X^+Y^-\rightarrow XY\\\\X^{2+}Y^-\rightarrow XY_2\\\\X^{3+}Y^-\rightarrow XY_3

- Y²⁻

X^+Y^{2-}\rightarrow X_2Y\\\\X^{2+}Y^{2-}\rightarrow X_2Y_2\rightarrow XY\\\\X^{3+}Y^{2-}\rightarrow X_2Y_3

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X^+Y^{3-}\rightarrow X_3Y\\\\X^{2+}Y^{3-}\rightarrow X_3Y_2 \\\\X^{3+}Y^{3-}\rightarrow X_3Y_3\rightarrow XY

Best regards!

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Answer:

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3 years ago
1. You have 19.7 grams of a material and wonder how many moles were formed. Your
Alex777 [14]

Answer:

No. If you were to set up what the friend said it would look like this:

19.7g × \frac{grams}{moles} ==>  \frac{19.7 grams}{1} ×

The above does not look right because the grams do not cancel out. If it's not cancelled, it would be included in your final answer, but you're looking for moles not grams.

When setting up a stoichiometry equation, you have to put the same unit of measurement (ex- grams, cm, mm, etc.) on the numerator of one side and the denominator of the other side.

This would cancel out the unit.

For example: If you wanted to find out how much of 320 cm are in a meter.

*there are 100 cm in 1 meter*

(1. always start with the given number!)

(2. set up the next fraction where you can cancel out cm)

\frac{320cm}{1} × \frac{1 meter}{100 cm} = 3.2 m

Going back, your friend would have to switch the units from grams/moles to moles/grams. It would cancel out grams, which would leave moles. Therefore, moles will be included in your final answer.

\frac{19.7 grams}{1} × \frac{moles}{grams} = _?_ moles

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2 years ago
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