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Natalka [10]
3 years ago
10

When the coffee is brewed according to directions, a pound of coffee beans yields 50 cups of coffee (4 cups = 1 qt). How many kg

of coffee are required to produce 250 cups of coffee?
Chemistry
2 answers:
Xelga [282]3 years ago
4 0

1 pound of coffee is said to yield 50 cups of coffee

So we need to find the pounds of coffee beans required to produce 250 cups

so if 50 cups need 1 pound of coffee beans

therefore 250 cups need - 1/50 x 250 = 5 pounds of coffee beans

since we have to find the amount of coffee beans needed in kg, we have to convert pounds to kg

1 pound = 0.45 kg

So if 1 pound is equivalent to 0.45 kg

then 5 pounds are equivalent to - 0.45 kg/pound x 5 pounds = 2.25 kg

therefore 2.25 kg of coffee beans are required

Gwar [14]3 years ago
3 0

\boxed{2.26796{\text{ kg}}} of coffee beans is required to produce 250 cups of coffee.

Further Explanation:

The standard that is used to compare the measurements is called unit. Units are of two types, basic or fundamental and derived units. The units that cannot number are called basic units. For example, mass, time and temperature are basic units. The units that depend upon basic units to express themselves are called derived units. Volume, area, and density are some examples of derived units.

One unit can be converted into another one with the help of proper conversion factor. Conversion factors are the ratios that are written in the form of fractions, which when multiplied by the original unit, gives the desired units.

It is given that one pound of coffee beans yields 50 cups of coffee. Therefore the amount of coffee beans required to produce 250 cups of coffee can be calculated as follows:

\begin{aligned}{\text{Amount of coffee beans required}}&=\left( {250{\text{ cups}}} \right)\left( {\frac{{1{\text{ pound}}}}{{50{\text{ cups}}}}} \right)\\&=5{\text{pounds}}\\\end{aligned}

The amount of coffee beans has to be converted into kilograms. The conversion factor for this is,

 1{\text{ pound}} = 0.453592{\text{ kg}}

Therefore the amount of coffee beans can be calculated as follows:

 \begin{aligned}{\text{Amount of coffee beans required}}&= \left( {{\text{5 pounds}}} \right)\left( {\frac{{{\text{0}}{\text{.453592 kg}}}}{{{\text{1 pound}}}}} \right)\\&= 2.26796{\text{ kg}}\\\end{aligned}

Learn more:

  1. What is the mass of 1 mole of viruses: brainly.com/question/8353774
  2. Determine the moles of water produced: brainly.com/question/1405182

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Basic concepts of chemistry

Keywords: unit, conversion factor, basic, fundamental, derived units, standard, mass, time, temperature, area, volume, density, coffee, coffee beans, 2.26796 kg, 5 pounds, 1 pound, 0.453592 kg.

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sattari [20]

Answer:

1) <u>16.8 L CO2</u>

<u>2) 36.96 L NH3</u>

<u>3) </u> <u>9.88 L CO2 </u>

<u>4) 56.99 L H2O</u>

Explanation:

How many liters of carbon dioxide gas will be produced when 75.0 g of calcium carbonate decomposes to form calcium oxide when at STP?

CaCO3 → CaO + CO2

Moles calcium carbonate = 75.0 grams / 100.09 g/mol

Moles calcium carbonate = 0.750 moles

For 1 mol CaCO3 we'll have 1 mol CaO and 1 mol CO2

For 0.750 moles CaCO3 we'll have 0.750 moles CO2

1 mol = 22.4 L

0.750 moles CO2 = 0.750 *22.4 L =<u> 16.8 L CO2</u>

2. Hydrogen gas reacts with 23.1 g of nitrogen gas to produce ammonia (NH3). What volume of ammonia will be produced at STP?

3H2 + N2 → 2NH3

Moles N2 = 23.1 grams / 28.0 g/mol

Moles N2 = 0.825 moles

For 3 moles H2 we need 1 mol N2 to produce 2 moles NH3

For 0.825 moles N2 we'll have 2*0.825 = 1.65 moles NH3

1 mol = 22.4 L

1.65 mol = 1.65 * 22.4 L = <u>36.96 L NH3</u>

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3. Iron (III) oxide reacts with carbon monoxide to form iron and carbon dioxide. How many liters of carbon dioxide will be produced from 23.5 g of iron (III) oxide when at STP?

Fe2O3 + 3CO → 2Fe + 3CO2

Moles Fe2O3 = 23.5 grams / 159.69 g/mol

Moles Fe2O3 = 0.147 moles

For 1 mol Fe2O3 we need 3 moles CO to produce 2 moles Fe and 3 moles CO2

For 0.147 moles Fe2O3 we'll have 3*0.147 = 0.441 moles CO2

1 mol = 22.4 L

0.441 moles = 22.4 * 0.441 = <u>9.88 L CO2 </u>

<u />

<u />

<u />

4.How many liters of water vapor would be produced in the combustion of 12.5L of ethane, C2H6 at STP?

2C2H6 + 7O2 →4CO2 + 6H2O

22.4 L = 1 mol

12.5 L = 0.848 moles C2H6

For 2 moles C2H6 we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

For 0.848 moles C2H6 we'll have 3*0.848 =  2.544 moles H2O

1 mol = 22.4 L

2.544 moles = 22.4 L * 2.544 = <u>56.99 L H2O</u>

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A 0.4021-g sample of a purified organic acid was dissolved in water and titrated potentiometrically. A plot of the data revealed
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Answer:

The molar mass of the acid is 167.5 g/mol

Explanation:

A 0.4021-g sample of a purified organic acid was dissolved in water and titrated potentiometrically. A plot of the data revealed a single end point after 19.31 mL of 0.1243 M NaOH had been introduced. Calculate the molecular mass of the acid.

Step 1: Data given

Mass of the sample of a purified organic acid = 0.4021 grams

Molarity = 0.1243 M

Volume needed to reach the end point = 19.1 mL = 0.01931 L

Step 2: Calculate the number of moles NaOH

Moles NaOH = molarity NaOH  * volume

Moles NaOH = 0.1243 M * 0.01931 L

Moles NaOH = 0.00240 moles

Step 3: Calculate moles of the acid

We'll need 0.00240 moles of acid to neutralize 0.00240 moles of NaOH ( it's a single end point)

Moles acid = 0.00240 moles

Step 4: Calculate molar mass of the acid

Molar mass = mass / moles

Molar mass = 0.4021 grams / 0.00240 moles

Molar mass = 167.5 g/mol

The molar mass of the acid is 167.5 g/mol

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