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2 years ago

Branches of chemistry with definition and examples in real life

Chemistry

1 answer:

nataly862011 [7]2 years ago

8 0

Explanation:The five major branches of chemistry are organic, inorganic, analytical, physical, and biochemistry.

...

Sub-branches of physical chemistry include:

Photochemistry — the study of the chemical changes caused by light.

Surface chemistry — the study of chemical reactions at surfaces of substances

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Should the United States continue to develop/use nuclear power? Why or why not?

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Answer:

Yes

Explanation:

because they should

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2 years ago

Write the name and molecular formula of an organic compounds having its name suffixed with -ol and having 2 carbon atoms in the

777dan777 [17]

Ethanol C₂H₆O

Explanation:

When ethanol (CH₃-CH₂-OH) is heated in the presence of the sulphuric acid (H₂SO₄) it will produce ethylene (CH₂=CH₂ ) and water (H₂O).

CH₃-CH₂-OH → CH₂=CH₂ + H₂O

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3 years ago

The equilibrium constant, Kc, for the following reaction is 0.967 at 650 K. 2NH3(g) N2(g) 3H2(g) When a sufficiently large sampl

AlekseyPX

Answer: Concentration of $NH_3$ in the equilibrium mixture is 0.31 M

Explanation:

Equilibrium concentration of $H_2$ = 0.729 M

The given balanced equilibrium reaction is,

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

Initial conc. x 0 0

At eqm. conc. (x-2y) M (y) M (3y) M

The expression for equilibrium constant for this reaction will be:

3y = 0.729 M

y = 0.243 M

$K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}$

Now put all the given values in this expression, we get :

$K_c=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}$

$0.967=\frac{0.243\times (0.729)^3}{(x-2\times 0.243)^2}$

$x=0.80$

concentration of $NH_3$ in the equilibrium mixture = $0.80-2\times 0.243=0.31$

Thus concentration of $NH_3$ in the equilibrium mixture is 0.31 M

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3 years ago

Plants and animals can cause weathering of rocks. True or false

Keith_Richards [23]

True hope this helps you.

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2 years ago

Read 2 more answers

400. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution

ch4aika [34]

<u>Answer:</u> The molecular weight of protein is $1.14\times 10^2g/mol$

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

or,

$\pi=i\times \frac{m_{solute}\times 1000}{M_{solute}\times V_{solution}\text{ (in mL)}}}\times RT$

where,

$\pi$ = Osmotic pressure of the solution = 0.0861 atm

i = Van't hoff factor = 1 (for non-electrolytes)

$m_{solute}$ = mass of protein = 400 mg = 0.4 g (Conversion factor: 1 g = 1000 mg)

$M_{solute}$ = molar mass of protein = ?

$V_{solution}$ = Volume of solution = 5.00 mL

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[25+273]K=298K$

Putting values in above equation, we get:

$0.0861atm=1\times \frac{0.4g\times 1000}{M\times 100}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\M=1136.62g/mol=1.14\times 10^2g/mol$

Hence, the molecular weight of protein is $1.14\times 10^2g/mol$

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3 years ago