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iris [78.8K]
3 years ago
13

it takes 513 kj to remove a mole of electrons from the atoms at the surface of a piece of metal. how much energy does it take to

remove a single electron from n atom at the surface of the metal
Chemistry
1 answer:
schepotkina [342]3 years ago
8 0

Answer:

The right solution is "8.5\times 10^{-19} \ joule".

Explanation:

As we know,

1 mole electron = 6.023\times 10^{23} \ no. \ of \ electrons

Total energy = 513 \ KJ

                     = 513\times 1000 \ joule

For single electron,

The amount of energy will be:

= \frac{513\times 1000}{(6.023\times 10^{23})}

= 8.5\times 10^{-19} \ joule

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A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s
Lilit [14]

Answer:

pH = 1.32

Explanation:

                 H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount  to buffer solutions.  In the first  deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g  =  0.074 mol H₂M

54 mL x  1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

           

3 0
3 years ago
Identify the products in the equation below (the “I” is Iodine—to differentiate capital “I” from lowercase “l”)
Inessa [10]

Answer : The products are Silver sulfide, (Ag_2S) and Sodium iodide, (NaI).

Explanation :

The given balanced chemical reaction is,

2AgI+Na_2S\rightarrow Ag_2S+2NaI

From the given balanced reaction, we conclude that the 2 moles of silver iodide react with the 1 mole of sodium sulfide to give product as 1 mole of silver sulfide and 2 moles of sodium iodide.

In a chemical reaction, reactants are represent on the left side of the right-arrow and products are represent on the right side of the right-arrow.

Therefore, in a chemical reaction the products are Silver sulfide and Sodium iodide.


5 0
3 years ago
Classify each of these reactions. Ba(ClO3)2 → BaCl2 3 O2 double replacement combination single replacement decomposition CaO CO2
faltersainse [42]

Answer:

Ba(ClO₃)₂ → BaCl₂ +  3 O₂ <em>Descomposition. </em>

CaO + CO₂ → CaCO₃ <em>Combination. </em>

NaNO₂ +  HCl → NaCl + HNO₂ <em>Double replacement. </em>

Mg + ZnSO₄ → MgSO₄ + Zn <em>Single replacement. </em>

Explanation:

A combination reaction is defined as a reaction in which two or more substances combine to form a single new substance.

A + B → AB

A descomposition reaction is defined as a reaction in which a compound breaks down into two or more simpler substances.

AB → A + B

A double replacement is a chemical reaction were the positive and negative ions of two ionic compounds exchange places to form two new compounds.

AB + CD → AD + CB

A single replacement is another type of reaction were one element replaces a similar element in a compound.

A + BC → AC + B

Thus, with this information it is possible to classify these reactions as:

Ba(ClO₃)₂ → BaCl₂ +  3 O₂ <em>Descomposition. </em>One single molecule breaks down into two or more molecules

CaO + CO₂ → CaCO₃ <em>Combination. </em>Two substances are combined to form one single molecule

NaNO₂ +  HCl → NaCl + HNO₂ <em>Double replacement. </em>Na and H are exchange places to form two new compounds

Mg + ZnSO₄ → MgSO₄ + Zn <em>Single replacement. </em>Mg is replacing Zn.

I hope it helps!

6 0
3 years ago
No spam or links! Thanks!
Monica [59]

Answer:

364 K or 91°C

Explanation:

Applying,

V₁/T₁ = V₂/T₂................ Equation 1

Where V₁ = Initial Volume, V₂ = Final volume, T₁ = initial Temperature, T₂ = final Temperature.

make T₂ the subject of the equation,

T₂ = V₂T₁/V₁................. Equation 2

From the question,

Given: V₁ = 375 mL, V₂ = 500 mL, T₁ = 0.0°C = (273+0) K = 273 K

Substitute these values into equation 2

T₂ = (500×273)/375

T₂ = 364 K

T₂ = (364-273) °C = 91 °C

5 0
3 years ago
You are a NASCAR pit crew member. Your employer is leading the race with 20 laps to go. He just finished a pit stop and has 5.0
jek_recluse [69]

Answer:

Since there are 3500 g of fuel left in the tank, and he needs only 1687.5 g to complete 20 laps, he has enough fuel to complete the race. I will tell the driver that he does not need to make another pit stop as he has enough fuel to complete the race.

Explanation:

Density = mass / volume

Density of fuel = 700 g/ 1 gal

Therefore, the mass of fuel in 1 gallon = 700 g

The driver has 5.0 gallons of fuel in the tank.

The mass of 5.0 gallons of fuel = 5 × 700 = 3500 g of fuel

Equation of the combustion of fuel, C₅H₁₂ is given below:

C₅H₁₂ + 8 O₂ ---> 6 H₂O + 5 CO₂

1 mole C₅H₁₂ requires 8 moles of O₂

1 mole of C₅H₁₂ has a mass = 72 g

8 moles of O₂ has a mass = 256 g

Therefore, 300 g of O₂ will require 300 × (72/256) g of C₅H₁₂ = 84.375 g of C₅H₁₂

84.375 g of fuel is used by the car per lap;

20 laps will require 20 × 84.375 g of fuel = 1687.5 g of fuel.

Since there are 3500 g of fuel left in the tank, and he needs only 1687.5 g to complete 20 laps, he has enough fuel to complete the race. I will tell the driver that he does not need to make another pit stop as he has enough fuel to complete the race.

7 0
3 years ago
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