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3 years ago

What type or process is involved in going from elements to compounds and vice versa

1 answer:

7 0

Mixtures to pure substances: separation processes
Pure substances to mixtures: mixing 

Both are physical changes, since most of those processes do not require chemical changes in the separation processes.

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How do synthesis reactions and a decomposition reactions differ

Minchanka [31]

A synthesis would be A+B-->AB its combining, decomposition is AB-->A+B, its the breaking up of a substance.

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3 years ago

Freshwater is distributed in both time and space

Neko [114]

Answer:

Unevenly

Explanation:

Fresh water is distributed unevenly in both time and space.

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2 years ago

Can someone help me please I will mark you brainliest

lara [203]

1: a, 2:a, 3:a, 4:c, 5:d, 6:d, 7:c;

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2 years ago

3. Give one example of an element and one

zhenek [66]

Answer:

Element: H (hydrogen)

Compound: H2O (water)

Explanation: An element is made up of a singular atom while a compound is made up of 2 or more

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3 years ago

Question List (4 items) (Drag and drop into the appropriate area) Find the volume of HCl that will neutralize the base. Find the

expeople1 [14]

The question is incomplete, the complete question is:

The solubility of slaked lime, $Ca(OH)_2$ , in water is 0.185 g/100 ml. You will need to calculate the volume of $2.50\times 10^{-3}$ M HCl needed to neutralize 14.5 mL of a saturated

Answer: The volume of HCl required is 290mL, the mass of $Ca(OH)_2$ is 0.0268g, the moles of

Explanation:

Given values:

Solubility of $Ca(OH)_2$ = 0.185 g/100 mL

Volume of $Ca(OH)_2$ = 14.5 mL

Using unitary method:

In 100 mL, the mass of $Ca(OH)_2$ present is 0.185 g

So, in 14.5mL. the mass of $Ca(OH)_2$ present will be = $\frac{0.185}{100}\times 14.5=0.0268g$

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of $Ca(OH)_2$ = 0.0268 g

Molar mass of $Ca(OH)_2$ = 74 g/mol

Plugging values in equation 1:

$\text{Moles of }Ca(OH)_2=\frac{0.0268g}{74g/mol}=0.000362 mol$

Moles of $OH^-$ present = $(2\times 0.000362)=0.000724mol$

The chemical equation for the neutralization of calcium hydroxide and HCl follows:

$Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O$

By the stoichiometry of the reaction:

Moles of $OH^-$ = Moles of $H^+$ = 0.000724 mol

The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}$ .....(2)

Moles of HCl = 0.000724 mol

Molarity of HCl = $2.50\times 10^{-3}$

Putting values in equation 2, we get:

$2.50\times 10^{-3}mol=\frac{0.000724\times 1000}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.000725\times 1000}{2.50\times 10^{-3}}=290mL$

Hence, the volume of HCl required is 290mL, the mass of $Ca(OH)_2$ is 0.0268g, the moles of

5 0

3 years ago