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xxMikexx [17]
3 years ago
5

Mass = 200.0 g volume = 100.0cm3 What is the density?

Chemistry
1 answer:
Marianna [84]3 years ago
3 0
The density of an object is defined as its mass divided by its volume. Mathematically, density = Mass / Volume. The unit of density is kilogram per cubic meter, kg / m^3 or g /cm^3.
For the question given above: the 
Mass = 200.0 g
Volume = 100.0 cm^3
Therefore, Density = Mass / Volume = 200 / 100  = 2
Thus,  the density of the object is 2 g /cm^3.
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Where must the moon be in its orbit for the full moon phase to be visible?
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3 years ago
The normal boiling point of ethanol (C2H5OH) is 78.3 oC and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the chan
Sloan [31]

Answer:

The  change in entropy in the system is -231.5 J/K

Explanation:

Step 1: Data given

The normal boiling point of ethanol (C2H5OH) is 78.3 °C = 351.45 K

The molar enthalpy of vaporization of ethanol is 38.56 kJ/mol = 38560 J/mol

Mass of ethanol = 97.2 grams

Pressure = 1 atm

Step 2: Calculate the entropy change of vaporization

The entropy change of vaporization = molar enthalpy of vaporization of ethanol / temperature

The entropy change of vaporization = 38560 J/mol / 351.45 K

The entropy change of vaporization = -109.72 J/k*mol

Step 3: Calculate moles of ethanol

Moles ethanol = mass / molar mass ethanol

Moles ethanol = 97.2 grams / 46.07 g/mol

Moles ethanol = 2.11 moles

Step 4: Calculate  change in entropy

For 1 mol = -109.72 J/K*mol

For 2.11 moles = -231.5 J/K

The  change in entropy in the system is -231.5 J/K

7 0
3 years ago
A what is the part that doesn't change during the experiment
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Explanation:

7 0
3 years ago
Read 2 more answers
What mass of HCL, in grams, is required to react with 0.610 g of al(oh)3 ?
kompoz [17]

Answer: 0.8541 grams of HCl will be required.

Explanation: Moles can be calculated by using the formula:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of Al(OH)_3 = 0.610 g

Molar mass of Al(OH)_3 = 78 g/mol

\text{Number of moles}=\frac{0.610g}{78g/mol}

Number of moles of Al(OH)_3 = 0.0078 moles

The reaction between Al(OH)_3 and HCl is a type of neutralization reaction because here acid and base are reacting to form an salt and also releases water.

Chemical equation for the above reaction follows:

Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O

By Stoichiometry,

1 mole of  Al(OH)_3 reacts with 3 moles of HCl

So, 0.0078 moles of Al(OH)_3 will react with \frac{3}{1}\times 0.0078 = 0.0234 moles

Mass of HCl is calculated by using the mole formula, we get

Molar mass of HCl = 36.5 g/mol

Putting values in the equation, we get

0.0234moles=\frac{\text{Given mass}}{36.5g/mol}

Mass of HCl required will be = 0.8541 grams

3 0
3 years ago
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