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3 years ago

Explain the difference between extensive properties and intensive properties.

Chemistry

1 answer:

Gnom [1K]3 years ago

5 0

Answer:

Extensive properties depend on de mass, intensive properties not.

Explanation:

The extensive properties depend on the mass and change with the addition or reducction of the mass.

The intensive properties dont depend on the mass. For example: temperature.

I hope I help you.

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How does a molecular formula differ from a structural formula?

mr Goodwill [35]

Explanation:

Molecular formulas show correct and accurate number of each type of the atoms which are present in molecule.

On the other hand, structural formulas show arrangement of atoms and covalent bonds between them.

For example,

The molecular formula for carbon dioxide is $CO_2$

The structural formula is O = C = O

7 0

3 years ago

Is boiling a chemical reaction?

swat32

Answer:

yes, because water is reacting with air and heat and creating a boiling effect.

Explanation: kracken doge

7 0

3 years ago

Read 2 more answers

Given the following balanced equation at 120°C: A(g) + B(g) ⇋ 2 C(g) + D(s)(a) At equilibrium a 4.0 liter container was found to

BlackZzzverrR [31]

Answer:

a) kc = 0,25

b) [A] = 0,41 M

c) [A] = 0,8 M

[B] =0,2 M

[C] = 0,2M

Explanation:

The equilibrium-constant expression is defined as the ratio of the concentration of products over concentration of reactants. Each concentration is raised to the power of their coefficient.

Also, pure solid and liquids are not included in the equilibrium-constant expression because they don't affect the concentration of chemicals in the equilibrium.

If global reaction is:

A(g) + B(g) ⇋ 2 C(g) + D(s)

The kc = $\frac{[C]^2}{[A][B]}$

a) The concentrations of each compound are:

[A] = $\frac{1,60 moles}{4,0 L}$ = 0,4 M

[B] = $\frac{0,40 moles}{4,0 L}$ = 0,1 M

[C] = $\frac{0,40 moles}{4,0 L}$ = 0,1 M

kc = $\frac{[0,1]^2}{[0,4][0,1]}$ = 0,25

b) The addition of B and D in the same amount will, in equilibrium, produce these changes:

[A] = $\frac{1,60-x moles}{4,0 L}$

[B] = $\frac{0,60-x moles}{4,0 L}$

[C] = $\frac{0,60+2x moles}{4,0 L}$

0,25 = $\frac{[0,60+2x]^2}{[1,60-x][0,60-x]}$

You will obtain

3,75x² +2,95x +0,12 = 0

Solving

x =-0,74363479081119 → No physical sense

x =-0,043031875855476

Thus, concentration of A is:

$\frac{1,60-(-0,04 moles)}{4,0 L}$ = 0,41 M

c) When volume is suddenly halved concentrations will be the concentrations in equilibrium over 2L:

[A] = $\frac{1,60 moles}{2,0 L}$ = 0,8 M

[B] = $\frac{0,40 moles}{2,0 L}$ = 0,2 M

[C] = $\frac{0,40 moles}{2,0 L}$ = 0,2M

I hope it helps!

8 0

4 years ago

Read 2 more answers

For the following compound, determine each element and if the element is a metal or nonmetal. Also determine if the compound has

Artyom0805 [142]

NaCl - also known as sodium chloride or table salt; an ionic bond; it’s a chemical compound, so both
SO2 - also known as sulfur dioxide; a covalent bond; nonmetal
CaO - also known as calcium oxide; an ionic bond; metal
HF - also known as hydrogen fluoride; polar covalent; metal
NO2 - also known as nitrogen dioxide; covalent; nonmetal
H2O - also known as water; covalent; nonmetal

4 0

3 years ago

Consider the reaction: 2 SO2(g)+O2(g)→2 SO3(g) If 285.5 mL of SO2 reacts with 158.9 mL of O2 (both measured at 315 K and 50.0 mm

Tanzania [10]

Answer:

a. Oxygen is the limiting reagent. $n_{SO_3}^{Theoretical}=8.096x10^{-4}mol SO_3$

b. $Y=58.9$ %

Explanation:

Hello,

a. Limiting reagent and sulfur trioxide's theoretical yield.

At first, we must compute the involved moles for both sulfur dioxide's and oxygen's as follows, considering the volumes in liters and the pressure in atm of 50.0mmHg*1atm/760mmHg=0.0658atm:

$n_{SO_2}=\frac{PV}{RT}=\frac{0.0658atm*0.2855L}{0.082\frac{atm*L}{mol*K}*315K} =7.273x10^{-3}molSO_2 \\n_{O_2}=\frac{PV}{RT}=\frac{0.0658atm*0.1589L}{0.082\frac{atm*L}{mol*K}*315K} =4.048x10^{-4}molO_2$

Afterwards, by considering the properly balanced chemical reaction:

$2SO_2(g)+O_2(g)-->2SO_3$

We compute the oxygen's moles that completely reacts with the previously computed $7.273x10^{-3}$ moles of $SO_2$ as follows:

$7.273x10^{-3}molSO_2*\frac{1molO_2}{2molSO_2} =3.6365x10^{-3}molO_2$

That result let us know that the oxygen is the limiting reagent since just $4.048x10^{-4}$ moles are available in comparison with the $3.6365x10^{-3}$ moles that completely would react with $7.273x10^{-3}$ moles of $SO_2$ .