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4 years ago

The graph shows that the reaction is at equilibrium after 60. seconds because the concentrations of both NO2(g) and N2O4(g) are

Chemistry

2 answers:

Monica [59]4 years ago

8 0

I think the reason why the reaction is at equilibrium after 60 seconds is because the concentration of both NO2 and N2O4 are : Constant, which means that they do not undergone any change

hope this helps

OlgaM077 [116]4 years ago

4 0

Answer: The correct answer is Option 3.

Explanation:

Chemical equilibrium is defined as the point where the concentrations of reactants and the products do not change with time.

The equation for the equilibrium between $NO_2\text{ and }N_2O_4$ follows:

$N_2O_4\rightleftharpoons NO_2$

Initially, the concentration of the reactants decrease and the concentration of the products increase as seen from the graph. But after 60 seconds, the concentration of both reactants and products do not change and hence equilibrium is attained.

Hence, the correct answer is Option 3.

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An oxide of nitrogen is 25.9% N by mass, has a molar mass of 108 g/mol, and contains no nitrogen-nitrogen or oxygen-oxygen bonds

Sonja [21]

Answer:

The compound is N2O4

Explanation:

We have certain important pieces of information about the compound;

1) it is an oxide (a binary compound of nitrogen and oxygen)

2) there are no N-N bonds present

3) there are no O-O bonds present

Since it contains only nitrogen and oxygen then nitrogen accounts for 25.9% of the molecule by mass then oxygen should account for (100-25.9) = 74.1% oxygen

Relative atomic mass of oxygen = 16

Relative atomic mass of nitrogen = 14

We now deduce the empirical formula

Nitrogen. Oxygen

25.9/14. 74.1/16

1.85/1.85. 4.6/1.85 (divide through by the lowest ratio)

1 2

Empirical formula is NO2

To find the molecular formula

(NO2)n = 108

(14+2(16))n= 108

46n=108

n= 108/46

n= 2

Therefore molecular formula= N2O4

4 0

3 years ago

Does the energy go from the surrounding to the chemicals or from the chemicals to the surround in an exothermic reaction?

Karo-lina-s [1.5K]

In general terms, exothermic reactions release energy, so the energy goes from the system to the surroundings.

4 0

3 years ago

A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of

zaharov [31]

Answer:

a) pH = 2.88

b) pH = 4.598

c) pH = 5.503

d) pH = 8.788

e) pH = 12.097

Explanation:

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.75 E-5 = [H3O+]*[CH3COO-] / [CH3COOH]

a) 0.0 mL KOH:

mass balance:

⇒ C CH3COOH = [CH3COOH] + [CH3COO-] = 0.100 M

charge balance:

⇒ [H3O+] = [CH3COO-]

⇒ 1.75 E-5 = [H3O+]²/(0.100 - [H3O+])

⇒ [H3O+]² + 1.75 E-5[H3O+] - 1.75 E-6 = 0

⇒ [H3O+] = 1.314 E.3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.88

b) 5.0 mL KOH:

CH3COOH + KOH ↔ CH3COONa + H2O

∴ C CH3COOH = ((0.025)(0.100) - (5 E-3)(0.200))/(0.025+5 E-3)

⇒ C CH3COOH = 0.05 M

∴ C KOH = ((5 E-3)(0,200))/(0.025+5 E-3) = 0.033 M

mass balance:

⇒ C CH3COOH + C KOH = [CH3COOH] + [CH3COO-] = 0.05 + 0.033 = 0.083 M

charge balance:

⇒ [H3O+] + [K+] = [CH3COO-]

⇒ [CH3COO-] = [H3O+] + 0.033

⇒ 1.75 E-5 = ([H3O+]*([H3O+] + 0.033))/(0.083 - ([H3O+] + 0.033))

⇒ 1.75 E-3 = ([H3O+]² + 0.033[H3O+])/(0.05 - [H3O+])

⇒ 8.75 E-7 - 1.75 E-5[H3O+] = [H3O+]² + 0.033[H3O+]

⇒ [H3O+]² +0.03302[H3O+] - 8.75 E-7 = 0

⇒ [H3O+] = 2.523 E-5 M

⇒ pH = 4.598

equivalent point:

(C*V)acid = (C*V)base

⇒ (0.100 M)*(0.025 L) = (0.200 M)( Vbase)

⇒ Vbase = 0.0125L = 12.5 mL

c) 10.0 mL KOH:

∴ C CH3COOH = 0.0143 M

∴ C KOH = 0.057 M

as in the previous point, starting from the mass and charge balances, we obtain:

⇒ [H3O+] = 3.1386 E-6 M

⇒ pH = 5.503

d) 12.5 mL KOH:

at the equivalence point, there is complete salt formation, then the pH is calculated through the salt:

CH3COO- + H2O ↔ CH3COOH - OH-

∴ Kw/Ka = 1 E-14/1.75 E-5 = 5.714 E-10 = [CH3COOH]*[OH-]/[CH3COO-]

∴ [CH3COO-] = (0.025)(0.100))/(0.025+0.0125) = 0.066 M

mass balance:

⇒ 0.066 = [CH3COOH] + [CH3COO-]..........(1)

charge balance:

⇒ [K+] = [OH-] + [CH3COO-] = 0.066 M.........(2)

∴ [K+] = C CH3COO- = 0.066 M

(1) = (2):

⇒ [OH-] = [CH3COOH].......(3)

⇒ 5.714 E-10 = [OH-]² / (0.066 - [OH-])

⇒ [OH-]² + 5.714 E-10[OH-] - 3.7712 E-11 = 0

⇒ [OH-] = 6.1408 e-6 m

⇒ pOH = 5.212

⇒ pH = 14 - pOH = 8.788

d) 15.0 mL KOH:

after the equivalence point there is salt and excess base (OH-); ph is calculated from excess base:

⇒ C KOH = ((0.015)(0.200) - (0.025)(0.100)) / (0.025 + 0.015) = 0.0125 M

⇒ [OH-] ≅ C KOH = 0.0125 M

⇒ pOH = 1.903

⇒ pH = 12.097

8 0

3 years ago

How many grams of hydrochloric acid will react completely with a block of gold is 3.2 cm by 3.8 cm by 2.8 cm, if the density of

JulijaS [17]

Answer:
mass of HCl = 243.5426 grams

Explanation:
1- we will get the mass of the reacting gold:
volume of gold = length * width * height
volume of gold = 3.2 * 3.8 * 2.8 = 34.048 cm^3 = 34.048 ml
density = mass / volume
Therefore:
mass = density * volume
mass of gold = 19.3 * 34.048 = 657.1264 grams

2- we will get the number of moles of the reacting gold:
number of moles = mass / molar mass
number of moles = 657.1264 / 196.96657
number of moles = 3.3362 moles

3- we will get the number of moles of the HCl:
First, we will balanced the given equation. The balanced equation will be as follows:
Au + 2HCl ......> AuCl2 + H2
This means that one mole of Au reacts with 2 moles of HCl.
Therefore 3.3362 moles will react with 2*3.3362 = 6.6724 moles of HCL

4- we will get the mass of the HCl:
From the periodic table:
molar mass of H = 1 gram
molar mass of Cl = 35.5 grams
Therefore:
molar mass of HCl = 1 + 35.5 = 36.5 grams/mole
number of moles = mass / molar mass
Therefore:
mass = number of moles * molar mass
mass of HCl = 6.6724 * 36.5
mass of HCl = 243.5426 grams

Hope this helps :)

4 0

3 years ago

An exothermic reaction has a positive enthalpy (heat) of reaction.(T/F)

mestny [16]

Answer:

True.

Explanation:

An exothermic reaction has a positive enthalpy (heat) of reaction. However, it can be negative in some circumstances.

5 0

3 years ago

Read 2 more answers