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2 years ago

What effect does mass have on an objects density?

Chemistry

1 answer:

xz_007 [3.2K]2 years ago

6 0

The density of an object or quantity of matter is its mass divided by its volume.

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Based on the following thermodynamic data, calculate the boiling point of ethanol in degrees Celsius. Substance ΔH∘f (kJ/mol) S∘

levacccp [35]

Answer:

76.03 °C.

Explanation:

Equation:

C2H5OH(l) --> C2H5OH(g)

ΔHvaporization = ΔH(products) - ΔH (reactants)

= (-235.1 kJ/mol) - (-277.7 kK/mol)

= 42.6 kJ/mol.

ΔSvaporization = ΔS(products) - ΔS(reactants)

= 282.6 J/K.mol - 160.6 J/K.mol

= 122 J/K.mol

= 0.122 kJ/K.mol

Using gibbs free energy equation,

ΔG = ΔH - TΔS

ΔG = 0,

T = ΔH/ΔS

T = 42.6/0.122

= 349.18 K.

Coverting Kelvin to °C,

= 349.18 - 273.15

= 76.03 °C.

8 0

2 years ago

Solid added to liquid and the solid floats down to bottom of container physical or chemical change?

Makovka662 [10]

Answer:

Physical change

Explanation:

4 0

2 years ago

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$

From the reaction, we conclude that

As, 2 mole of $Na$ react to give 2 mole of $NaCl$

So, 0.217 mole of $HCl$ react to give 0.217 mole of $NaCl$

Now we have to calculate the mass of $NaCl$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

Molar mass of $NaCl$ = 58.5 g/mole

$\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g$

Now we have to calculate the percent yield of this reaction.

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = $\frac{5.24g}{12.7g}\times 100$

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

4 0

2 years ago

2. What is the name of the compound AIPO?

Andru [333]

Answer:

Aluminum phosphate

Explanation:

7 0

1 year ago

A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution? 1.0 × 10–5 M OH– 1.0 ×

Triss [41]

Answer:

1.0 × 10⁻⁹ M.

Explanation:

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

[H₃O⁺] = 1.0 x 10⁻⁵ M.

∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴/(1.0 x 10⁻⁵ M) = 1.0 × 10⁻⁹ M.

6 0

2 years ago

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