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Shtirlitz [24]
3 years ago
14

List the phase changes that result of an increase of molecular kinetic energy pls help!! 30 points!

Chemistry
1 answer:
Nina [5.8K]3 years ago
8 0

Answer:

Explanation:

The term change of phase means the same thing as the term change of state. ... i.e. during phase change, the energy supplied is used only to separate the molecules ; no part of it is used to increase the kinetic energy of the molecules. So its temperature will not rise, since kinetic energy of molecules remains the same.

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A- is a weak base. which equilibrium corresponds to the equilibrium constant ka for ha?
vekshin1

Answer:

K_a=\frac{[H^+][A^-]}{[HA]}

Explanation:

ka is defined as the dissociation constant of an acid. It is defined as the ratio of concentration of products to the concentration of reactants.

For the dissociation of weak acid, the chemical equation follows:

HA\rightleftharpoons H^++A^-

The equilibrium constant is defined by the equilibrium concentration of products over reactants:

K_a=\frac{[H^+][A^-]}{[HA]}

5 0
3 years ago
What is the net cell reaction for the cobalt-silver voltaic cell? express your answer as a chemical equation?
slava [35]

Answer: The net ionic equation for the cobalt-silver voltaic cell is Co(s)+3Ag^+(aq.)\rightarrow Co^{3+}(aq.)+3Ag(s)

Explanation: In cobalt-silver voltaic cell, one half of the cell consists of cobalt electrode immersed in Co(NO_3)_3 solution ( which means that Co^{3+} are present in the solution) and other half of the cell consists of the Ag electrode immersed in AgNO_3 solution ( which means that Ag^+ is present in the solution)

The two electrodes are joined by the copper wire. The cobalt electrode acts as an anode and the silver electrode acts a  cathode.

At anode, oxidation reaction takes place and at cathode, reduction reaction takes place.

At Anode :                    Co(s)\rightarrow Co^{3+}(aq.)+3e^-

At Cathode:                   [Ag^+(aq.)+e^-\rightarrow Ag(s)]\times 3

Net ionic equation:   Co(s)+3Ag^+(aq.)\rightarrow Co^{3+}(aq.)+3Ag(s)

6 0
3 years ago
What is the pH when 10.0 mL of 0.20 M potassium hydroxide is added to 30.0 mL of 0.10 M cinnamic acid, HC9H7O2 (Ka = 3.6 × 10–5)
astra-53 [7]

Answer:-

Solution:- As is clear from the given Ka value, Cinnamic acid is a weak acid. let's calculate the moles of acid and KOH added to it from their given molarities and mL.

For KOH,  10.0mL(\frac{1L}{1000mL})(\frac{0.20mol}{1L})

= 0.002 mol

For Cinnamic acid,  30.0mL(\frac{1L}{1000mL})(\frac{0.10mol}{1L})

= 0.003 mol

Acid and base react as:

HC_9H_7O_2(aq)+KOH(aq)\rightleftharpoons KC_9H_7O_2(aq)+H_2O(l)

The reaction takes place in 1:1 mol ratio. Since the moles of acid are in excess, the acid is still remaining when all the kOH is used.

0.002 moles of KOH react with 0.002 moles of Cinnamic acid to form 0.002 moles of potassium cinnamate. Excess moles of Cinnamic acid = 0.003 - 0.002 = 0.001

As the solution have weak acid and it's salt(or we could say conjugate base), it is a buffer solution and the pH of the buffer solution could easily be calculated using Handerson equation:

pH=pKa+log(\frac{base}{acid})

pKa could be caluted from given Ka value using the formula:

pKa = - log Ka

pKa=-log3.6*10^-^5

pKa = 4.44

let's plug in the values in Handerson equation and calculate the pH:

pH=4.44+log(\frac{0.002}{0.001})

pH = 4.44+0.30

pH = 4.74

So, the first choice is correct, pH is 4.74.

6 0
3 years ago
How much sodium bicarbonate to raise alkalinity in pool.
almond37 [142]

Answer:

A rule of thumb is that 1.5 lbs. of baking soda per 10,000 gallons of water will raise alkalinity by about 10 ppm. If your pool's pH is tested below 7.2, add 3-4 pounds of baking soda. If you're new to adding pool chemicals, start by adding only one-half or three-fourths of the recommended amount.

4 0
2 years ago
A glass cup of orange juice is found to have pOH of 11.40. Calculate the concentration of the hydrogen ions in the juice.​
fgiga [73]

Answer: The concentration of hydrogen ion is 2.5 x 10∧-3.

Explanation:

It is well known that; pH + pOH = 14

∴ pH + 11.40 = 14

  pH = 14 -11.40 = 2.60

Remember that,

pH = - Log [ H+ ]

2.60 = - log [H+]

To get the  hydrogen ion concentration, we take the anti-log of 2.60.

[H+] = Antilog 2.60 = 2.5 x 10∧-3.

4 0
3 years ago
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