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2 years ago

13

A grocery store has 75 dozen packages of hotdogs In stock. If there are 10 hot dogs in each package, how many moles of hot dogs

is this?

2 answers:

Viktor [21]2 years ago

6 0

you can just divide 75 and 10

Zepler [3.9K]2 years ago

6 0

Answer:

$1.49*10^{-20}hotdogmoles$

Explanation:

First we will calculate the amount of hot dog units we have:

-75 dozen packages (a dozen is 12 )=75x12=900 packages

-each package has 10 hot dogs=900x10=9000 hot dog units

We know a mole of something is $6.022*10^{23}$ units of that something. And that will be our conversion factor for this excercise. We will convert now from 9000 units of hot dogs to moles of hot dogs like this: $9000 hotdogunits*\frac{1 hotdogmoles }{6.022*10^{23}hotdogunits }=1.49*10^{-20}hotdogmoles$

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Convert 7.8 liters of carbon tetra fluoride cfa to grams

dolphi86 [110]

To answer the question, we assume that the given compound is an ideal gas that at STP, one mole of the substance will occupy 22.4 L. From the given volume, we determine the number of moles of substance.
7.8 L / (22.4 L /mole) = 0.3482 moles of cfa
Then, we multiply this number of moles by the molar weight of cfa which is equal to 88 g/mol.
Multiplying,
weight = (0.3482 moles of cfa) x (88 g/mol) = <em>30.64 grams</em>

4 0

2 years ago

Let's say we have 2 Zinc atoms and 2 Sulfur atoms on one side of the equation. According to the Law of Conversation and Mass, ho

Viefleur [7K]

Answer:

Two

Explanation:

The law of conservation of mass states that in an isolated system the mass present is neither destroyed nor created by chemical changes or physical changes.

This tells us that the mass of reactants must be equal to the product mass.

If 1 atom of Zn react with one atoms of sulfur, the product will be 1 molecule of zinc sulfide according to the equation below

Zn(s) + S(s) ⇒ ZnS(s)

therefore two atoms each f zinc and sulfur will product two molecules of Zn sulfide

5 0

1 year ago

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

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2 years ago

Which of the following does not involve colligative properties?

fomenos

Colligative properties are usually used in relation to solutions.
Colligative properties are those properties of solutions, which depend on the concentration of the solutes [molecules, ions, etc.] in the solutions and not on the chemical nature of those chemical species. Examples of colligative properties include: vapour pressure depression, boiling point elevation, osmotic pressure, freezing point depression, etc.
For the question given above, the correct option is D. This is because the statement is talking about freezing point elevation, which is not part of colligative properties.

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2 years ago

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Choose the correct description for each of the

Aliun [14]

Answer:

Density: Physical Property

Flammability: Chemical Property

Solubility In Water: Physical Property

Reactivity With Water: Chemical Property

Melting Pot: Physical Property

Color: Physical Property

Odor: Physical Property

Explanation:

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2 years ago

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