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4 years ago

How does a system at equilibrium respond to the addition of more reactant

Chemistry

1 answer:

grin007 [14]4 years ago

7 0

Answer:

It favors the forward reaction.

Explanation:

According to Le Chatelier's Principle, when a system at equilibrium suffers a perturbation, the system will react in order to counteract the effect of such perturbation.

If more reactant is added, the system will try to decrease its concentration. It will do so by favoring the forward reaction, decreasing the concentration of the reactant and increasing the concentration of the products, in order to re-establish the equilibrium.

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The form of carbon created by plants is

Mars2501 [29]

Carbon dioxide (CO2)
in the process of respiration, oxygen and glucose react to form carbon dioxide and water.

8 0

3 years ago

Consider the reaction: CH4 + 2O2 = CO2 + 2H2O

crimeas [40]

Answer:

The answer to your question is:

a) 80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction CH4 + 2O2 ⇒ CO2 + 2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

MW CH4 = 16 g

MW O2 = 32 g

16 g of CH4 ---------------- 2(32) g of O2

20 g -------------- x

x = (20 x 64) / 16 = 80 g of O2

b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .

CH4 + 2O2 ⇒ CO2 + 2H2O

15 g 22 g

16 g of CH4 ---------------- 64 g of O2

15 g of CH4 --------------- x

x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

CH4 + 2O2 ⇒ CO2 + 2H2O

64 g of O2 ------------------ 44 g of CO2

22 g of O2 ------------------ x

x = (22 x 44)/ 64 = 15. 13 g of CO2

c. For the reaction CH4 + 2O2 = CO2 + 2H2O, if you have 10.31 g. of CH4 and an unknown amount of oxygen, and form 20.00 g. of CO2, i. Identify if there is a limiting reactant ______________ ii. Calculate the number of grams of the limiting reactant present if there is one. ______________

CH4 + 2O2 ⇒ CO2 + 2H2O

10.31 g 20 g

We can identify the limiting reactant if we know the quantity of the reactants, if we only know the quantity of one it is not posible to which is the limiting reactant.

4 0

3 years ago

Which ion does not have a noble gas configuration in its ground state?

goldfiish [28.3K]

I think the answer would be Ga3+

3 0

3 years ago

Calculate the molality of a 20.0% by mass ammonium sulfate (nh4)2so4 solution. the density of the solution is 1.117 g/ml.

olasank [31]

Hello!

We have the following data:

m1 (solute mass) = 20 % m/m
M1 (Molar mass of solute) (NH4)2 SO4 = ?
m2 (mass of the solvent) = ? (in Kg)

First we find the solute mass (m1), knowing that:

20% m/m = 20g/100mL

20 ------ 100 mL (0,1 L)
y g --------------- 1 L

y = 20/0,1
y = 200 g --> m1 = 200 g

Let's find Solute's Molar Mass, let's see:

M1 of (Nh4)2SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

We must find the volume of the solvent and therefore its mass (m2), let us see:

d = 1,117 g/mL
m = 200 g
v (volumen of solute) = ?

$d = \dfrac{m}{V} \to V = \dfrac{m}{d}$

$V = \dfrac{200\:\diagup\!\!\!\!g}{1,117\:\diagup\!\!\!\!g/mL} \to V = 179\:mL\:(volumen\:of\:solute)$

The solvent volume will be:

1000 -179 => V = 821 mL (volumen of disolvent)

If: 1 mL = 1g

Then the mass of the solvent is:

m2 (mass of the solvent) = 821 g → m2 (mass of the solvent) = 0,821 Kg

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{m_1}{M_1*m_2}$

$\omega = \dfrac{200}{132*0,821}$

$\omega = \dfrac{200}{108,372}$

$\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

_________________________________
_________________________________

Another way to find the answer:

We have the following data:

W (molality) = ? (in molal)
n (number of mols) = ?
m1 (solute mass) = 20 % m/m = 20g/100mL → (in g to 1L) = 200 g
m2 (disolvent mass) the remaining percentage, in the case: 80 % m/m = 800 g → m2 (disolvent mass) = 0,8 Kg
M1 (Molar mass of solute) (NH4)2 SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

Let's find the number of mols (n), let's see:

 $n = \dfrac{m_1}{M_1}$

$n = \dfrac{200}{132}$

$n \approx 1,5\:mol$

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{n}{m_2}$

$\omega = \dfrac{1,5}{0,8}$

$
\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

I hope this helps. =)

7 0

3 years ago

In which of the following states is nitrogen found at room temperature?

mamaluj [8]

At room temperature two atoms of nitrogen combine to form a colorless and odorless diatomic gas. it is found in gaseous state.

hope the answer is helpful.

4 0

3 years ago