Which type of chemical reaction occurs in C6H12 + 9O2 ----- 6CO2 + 6H2O?

blondinia [14]

Answer:

$[CO]=[Cl_2]=0.01436M$

$[COCl_2]=0.00064M$

Explanation:

Hello there!

In this case, according to the given chemical reaction at equilibrium, we can set up the equilibrium expression as follows:

$K=\frac{[CO][Cl_2]}{[COCl_2]}$

Which can be written in terms of x, according to the ICE table:

$0.32=\frac{x^2}{0.015M-x}$

Thus, we solve for x to obtain that it has a value of 0.01436 M and therefore, the concentrations at equilibrium turn out to be:

$[CO]=[Cl_2]=0.01436M$

$[COCl_2]=0.015M-0.01436M=0.00064M$

Regards!

5 0

3 years ago

Why do a car's tires need more air in the winter than they do in the summer?

Hunter-Best [27]

Answer: Option (d) is the correct answer.

Explanation:

In winter's, temperature of atmosphere is low and due to this molecules of air present in the tire come closer to each other as they gain potential energy and loses kinetic energy.

Hence, air pressure decreases and there is need to fill more air in the tire.

Whereas is summer's, temperature is high so, molecules of air inside the tire gain kinetic energy and move rapidly from one place to another due to number of collisions. So, air pressure increases and there is no need to fill more air inside the tire.

Thus, we can conclude that the temperature is lower, so the air inside the tires contracts.

3 0

3 years ago

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What is the pH of a 3.40 mM of acetic acid, 500 mL in which 50 mL of 1.00 M NaOAc has been added? Ka HOAc is 1.8 x 10-5? What pe

Deffense [45]

Answer:

a) pH = 4.213

b) % dis = 2 %

Explanation:

Ch3COONa → CH3COO- + Na+

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]

mass balance:

⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]

<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)

∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL

⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water

⇒ [ CH3COO- ] = [ H3O+ ] + 1.00

⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5

⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0

⇒ [ H3O+ ] = 6.12 E-5 M

⇒ pH = - Log [ H3O+ ] = 4.213

b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4

∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol

⇒ % dis = 3.4 / 1.7 = 2 %

7 0

3 years ago

What volume (L) of 0.250 M HNO3 is required to neutralize a solution prepared by dissolving 17.5 g of NaOH in 350 mL of water ?

IRINA_888 [86]

<h3>Answer:</h3>

1.75 L HNO₃

<h3>Explanation:</h3>

We are given;

Molarity of HNO₃ as 0.250 M

Mass of NaOH as 17.5 g

Volume of NaOH = 350 mL

We are required to calculate the volume of 0.250 M

We are going to first write the balanced reaction:

NaOH(aq) + HNO₃(aq) → NaNO₃(aq) + H₂O(l)

Then, we calculate the number of moles of NaOH

Moles = Mass ÷ Molar mass

Molar mass of NaOH = 39.997 g/mol

= 17.5 g ÷ 39.997 g/mol

= 0.4375 moles

We can now calculate the number of moles of HNO₃ using the mole ratio from the equation;

Mole ratio of NaOH to HNO₃ is 1 : 1

Therefore, if moles of NaOH are 0.4375 moles then;

Moles of HNO₃ will also be 0.4375 moles

We can now calculate the volume of HNO₃

Morality = Number of moles ÷ Volume

Thus;

Volume = Number of moles ÷ Molarity

= 0.4375 moles ÷ 0.250 M

= 1.75 L

Therefore, the volume of HNO₃ is 1.75 L

5 0

3 years ago

1. What is the wave speed of a wave that has a frequency of 100 Hz and a wavelength of 0.30 m?

Sav [38]

Wave speed = frequency x wavelength
= 100 x 0.3
= 30 m/s.

Hope this helps!

5 0

3 years ago

Read 2 more answers