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3 years ago

What is the density of a mineral with a mass of 41.2 g and a volume 8.2 cm3?

Chemistry

2 answers:

Tanya [424]3 years ago

7 0

Answer : The density of a mineral is, 5.024 g/ml

Solution : Given,

Mass of mineral = 41.2 g

Volume of mineral = $8.2cm^3=8.2ml$ $(1cm^3=1ml)$

Formula used :

$Density=\frac{Mass}{Volume}$

Now put all the given values in this formula, we get the density of mineral.

$Density=\frac{41.2g}{8.2ml}=5.02g/ml$

Therefore, the density of a mineral is, 5.024 g/ml

dezoksy [38]3 years ago

4 0

Hey there!:

mass = 41.2 g

Volume = 8.2 cm³

Therefore:

D = m / V

D = 41.2 / 8.2

D = 5.02 g/cm³

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How many grams are in 5.2 moles of Li2SO4

skelet666 [1.2K]

Answer:

572 g

Explanation:

Molar mass is the mass of 1 mol of an element or compound

molar mass of Li₂SO₄ is the sum of the products of the molar masses of the elements by the number of atoms in the compound

molar masses of each element making up lithium sulphate

Li - 7 g/mol

S - 32 g/mol

O - 16 g/mol

molar mass of Li₂SO₄ - (7 g/mol x 2) + ( 32 g/mol x 1) + ( 16 g/mol x 4 )

molar mass = 110 g/mol

mass of 1 mol of Li₂SO₄ is 110 g

therefore mass of 5.2 mol of Li₂SO₄ is - 110 g/mol x 5.2 mol = 572 g

mass is 572 g

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3 years ago

There are two isotopes of chlorine. below is a list of each and their mass and abundance. calculate the atomic mass of chlorine.

Vesna [10]

Chlorine is very lequid structure

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3 years ago

The specific heat of copper metal is 0. 385 J/(g °C). How much energy must be added to a 35. 0-gram sample of copper to change t

Rus_ich [418]

The amount of heat required for changing the temperature of copper has been 606 J. Thus, option B is correct.

Specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius.

The heat required to raise the temperature has been expressed as:

$\rm Heat=mass\;\times\;specific\;heat\;\times\;Change\;in\;temperature$

<h3>Computation for the heat energy required</h3>

The given specific heat of copper has been $\rm 0.385\;J/g^\circ C$

The mass of copper has been, $\rm 35\;g$

The initial temperature of copper has been, $\rm 20^\circ C$

The final temperature of copper has been, $\rm 65^\circ C$

The change in temperature has been, $\Delta T$

$\Delta T=\text{Final\;temperature-Initial\;temperature}\\\Delta T =65^\circ \text C-20^\circ \text C\\\Delta T=45^\circ \text C$

Substituting the values for the heat required as:

$\rm Heat=35\;g\;\times\;0.385\;J/g^\circ C\;\times\;45^\circ C\\Heat=606\;J$

The amount of heat required for changing the temperature of copper has been 606 J. Thus, option B is correct.

Learn more about specific heat, here:

brainly.com/question/2094845

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2 years ago

What would the products of a double-replacement reaction between KBr and CaO be? (Remember: In double-replacement reactions, the

11111nata11111 [884]

ANSWER

$\begin{gathered} \text{ 2KBr }+\text{ CaO }\rightarrow\text{ K}_2O\text{ }+\text{ CaBr}_2 \\ Option\text{ B} \end{gathered}$

EXPLANATION

Given that;

The two reactants are KBr and CaO

Double replacement reaction is a type of chemical reaction that occur when two reactants exchange cations and anions to yield new products.

$\text{ 2KBr + CaO }\rightarrow\text{ K}_2O\text{ + CaBr}_2$

Therefore, the resulting products of the given data are K2O + CaBr2

The correct answer is option B

4 0

1 year ago

A mixture contains only nacl and al2(so4)3. a 1.45 g sample of the mixture is dissolved in water, and an excess of naoh is added

Aleks [24]

When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced. The balanced chemical reaction is represented as-

Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄

On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.

As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.

So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 = $\frac{1.085 X 100}{1.45}$ =74.8 %.

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2 years ago