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inessss [21]
3 years ago
13

A student was trying to obtain lithium by electrolysis of aqueous lithium chloride. Was he successful?

Chemistry
1 answer:
Wittaler [7]3 years ago
6 0

The student was not successful.

Consider the standard reduction potentials.

Li⁺ + e⁻ ⇌ Li;                       E° = -3.04 V

2H₂O + 2e⁻ ⇌ H₂ + 2OH⁻; E° = -0.83 V

To reduce Li⁺ to Li, the student must apply 3.04 V.

However, it takes only 0.83 V to reduce water to hydrogen.

Thus, the student will get H₂ instead of Li.

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If 12.5 grams of the original sample of cesium-137 remained after 90.6 years, what was the mass of the original sample?
myrzilka [38]

Answer:

Mass of original sample = 100 g

Explanation:

Half life of cesium-137 = 30.17 years

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac{\ln2}{t_{1/2}}

k=\frac{\ln2}{30.17}\ year^{-1}

The rate constant, k = 0.02297 year⁻¹

Time = 90.6 years

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Initial concentration [A_0] = ?

Final concentration [A_t] = 12.5 grams

Applying in the above equation, we get that:-

12.5\ g=[A_0]e^{-0.02297\times 90.6}

[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g

<u>Mass of original sample = 100 g</u>

8 0
3 years ago
Sodium metal (Na) reacts explosively with water to form sodium hydroxide (NaOH) and hydrogen gas. How many moles of sodium are n
Sati [7]

Answer:

308 moles of sodium

Explanation:

The balanced equation for the chemical reaction between sodium metal (Na) and water (H₂O) is the following:

2 Na(s) + 2 H₂O → 2 NaOH(aq) + H₂(g)

From the equation, we can see that 2 moles of Na react with 2 moles of H₂O to give 2 moles of NaOH and 1 mol of H₂ (hydrogen gas). So the stoichiometric mole ratio between Na and H₂ is: 2 mol Na/1 mol H₂. Thus, we multiply the mole ratio by the moles of H₂ to be produced to obtain the moles of Na required:

moles of Na required =  2 mol Na/1 mol H₂ x 154 moles H₂ = 308 moles Na

Therefore, 308 moles of sodium are needed to produce 154 moles of hydrogen gas.

7 0
3 years ago
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6 0
2 years ago
Explain electronegativity and how it relates to both molecular and ionic bonding.
Keith_Richards [23]

Answer:

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3 0
2 years ago
In a neutralization reaction, 24.6 mL of 0.300 M H2SO4(aq) reacts completely with 20.0 mL of NaOH(aq). The products are Na2SO4(a
stiv31 [10]

Answer : The concentration of the NaOH solution is, 0.738 M

Explanation :

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.300M\\V_1=24.6mL\\n_2=1\\M_2=?\\V_2=20.0mL

Putting values in above equation, we get:

2\times 0.300M\times 24.6mL=1\times M_2\times 20.0mL

M_2=0.738M

Thus, the concentration of the NaOH solution is, 0.738 M

7 0
3 years ago
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