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2 years ago

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A bar of Sam's coconut oil soap has a mass of 132 grams. The rectangular bar is 8 centimeters long, 6 centimeters wide, and 2.5

centimeters high. How many grams per cubic centimeter ia there? Calculate the density of Sam's coconut oil soap.
Write your answer as a decimal.

1 answer:

Sophie [7]2 years ago

7 0

Answer:

1.1

Explanation:

density= 1.1g/cm3 there are 1.1 grams per cubic cm

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A compound is made up of 28 g N, 24 g C, 48 g O, and 8 g H .What is the empirical formula?

vovikov84 [41]

Answer:

$\rm C_2H_8N_2O_3$ .

Explanation:

<h3>Step One: calculate the coefficients. </h3>

Look up the relative atomic mass of these four elements on a modern periodic table:

$\rm C$ : approximately $12$ .
$\rm H$ : approximately $1$ .
$\rm N$ : approximately $14$ .
$\rm O$ : approximately $16$ .

The relative atomic mass of an element is numerically equal to the mass (in grams, $\rm g$ ,) of one mole of atoms of this element.

For example, the relative atomic mass of $\rm C$ is approximately $12$ . Therefore, each mole of $\rm C\!$ atoms would have a mass of $12\; \rm g$ .

This sample contains $24\; \rm g$ of carbon. That would correspond to approximately $\displaystyle \left(\frac{24}{12}\right)\; \rm mol = 2\; \rm mol$ of $\rm C$ atoms.

Similarly, for the other three elements:

$\displaystyle n(\mathrm{H}) \approx \frac{8\; \rm g}{1\; \rm g \cdot mol^{-1}} = 8\; \rm mol$ .

$\displaystyle n(\mathrm{N}) \approx \frac{28\; \rm g}{14\; \rm g \cdot mol^{-1}} = 2\; \rm mol$ .

$\displaystyle n(\mathrm{O}) \approx \frac{48\; \rm g}{16\; \rm g \cdot mol^{-1}} = 3\; \rm mol$ .

Hence, the ratio between these elements in this compound would be:

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ .

In the empirical formula of a compound, the coefficients should represent the smallest possible integer ratio between the number of atoms of these elements.

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ is indeed the smallest possible integer ratio between the number of atoms of these elements.

<h3>Step Two: arrange the elements in an appropriate order</h3>

Apply the Hill System to arrange these four elements in the empirical formula. In the Hill System:

If carbon, $\rm C$ , is present in this compound, then:

$\rm C$ (carbon) and then $\rm H$ (hydrogen) will be the first two elements listed in the formula (ignore the hydrogen if it is not in the compound.)
The other elements in this compound will be listed in alphabetical order.

If there is no carbon $\rm C$ in this compound, then list all the elements in this compound in alphabetical order.

Both $\rm C$ (carbon) and $\rm H$ (hydrogen) are found in this compound. Therefore, the first element in the list would be $\rm C\!$ . The second would be $\rm H\!$ , followed by $\rm N\!$ and then $\rm O\!$ .

Hence, the empirical formula of this compound would be $\rm C_2H_8N_2O_3$ .

6 0

2 years ago

When metals combine with nonmetals, the metallic atoms tend to 1. lose electrons and become positive ions 2. lose electrons and

Vika [28.1K]

The correct answer is 1. Lose electrons and become positive ions.

I hope my answer was beneficial to you! c:

5 0

2 years ago

Read 2 more answers

Please help me double-check my answer: Calculate the molarity of an aqueous solution that contains 36.5g KMnO4 and has a total v

Helen [10]

Answer:

The answer to your question is Molarity = 0.6158, I got the same answer as you.

Explanation:

Data

Molarity = ?

Mass of KMnO₄ = 36.5 g

Total volume = 375 ml

Process

1.- Calculate the Molar mass of KMnO₄

KMnO₄ = (1 x 39.10) + (54.94 x 1) + (16 x 4)

= 39.10 + 54.94 + 64

= 158.04 g

2.- Calculate the moles of KMnO₄

158.04 g of KMnO₄ ------------------- 1 mol

36.5 g of KMnO₄ --------------------- x

x = (36.5 x 1) / 158.04

x = 0.231 mol

3.- Convert the volume to liters

1000 ml -------------------- 1 L

375 ml --------------------- x

x = (375 x 1)/1000

x = 0.375 L

4.- Calculate the Molarity

Molarity = moles / volume

-Substitution

Molarity = 0.231 moles / 0.375 L

Result

Molarity = 0.6158

6 0

2 years ago

Read 2 more answers

Fill in the blanks with the word that best completes each statement. Scientists develop knowledge by making ------------about th

kotegsom [21]

Answer:

1. observations

2. hypothesis

3. experimentation

Explanation:

Observations are made, Hypothesies are formed from observations, and experimnets help alter and comfirm hypothesis.

3 0

2 years ago

When 13.6 g of calcium chloride, CaCl2, was dissolved in 100.0 mL of water in a coffee cup calorimeter, the temperature rose fro

DanielleElmas [232]

Answer:

THE ENTHALPY OF SOLUTION IS 3153.43 J/MOL OR 3.15 KJ/MOL.

Explanation:

1. write out the variables given:

Mass of Calcium chloride = 13.6 g

Change in temperature = 31.75°C - 25.00°C = 6.75 °C

Density of the solution = 1.000 g/mL

Volume = 100.0 mL = 100.0 mL

Specific heat of water = 4.184 J/g °C

Mass of the water = unknown

2. calculate the mass of waterinvolved:

We must first calculate the mass of water in the bomb calorimeter

Mass = density * volume

Mass = 1.000 * 100

Mass = 0.01 g

3. calculate the quantity of heat evolved:

Next is to calculate the quantity of heat evolved from the reaction

Heat = mass * specific heat of water * change in temperature

Heat = mass of water * specific heat *change in temperature

Heat = 13.6 g * 4.184 * 6.75

Heat = 13.6 g * 4.184 J/g °C * 6.75 °C

Heat = 384.09 J

Hence, 384.09J is the quantity of heat involved in the reaction of 13.6 g of calcium chloride in the calorimeter.

4. calculate the molar mass of CaCl2:

Next is to calculate the molar mas of CaCl2

Molar mass = ( 40 + 35.5 *2) = 111 g/mol

The number of moles of 13.6 g of CaCl2 is then:

Number of moles of CaCl2 = mass / molar mass

Number of moles = 13.6 g / 111 g/mol

Number of moles = 0.1225 mol

So 384.09 J of heat was involved in the reaction of 1.6 g of CaCl2 in a calorimter which translates to 0.1225 mol of CaCl2..

5. Calculate the enthalpy of solution in kJ/mol:

If 1 mole of CaCl2 is involved, the heat evolved is therefore:

Heat per mole = 384.09 J / 0.1225 mol

Heat = 3 135.43 J/mol

The enthalpy of solution is therefore 3153.43 J/mol or 3.15 kJ/mol.

5 0

2 years ago