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iris [78.8K]
2 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]2 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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In the manufacture of paper, logs are cut into small chips, which are stirred into an alkaline solution that dissolves several o
Kitty [74]

Answer:

The estimated feed rate of logs is 14.3 logs/min.

Explanation:

The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.

That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.

Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.

To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.

V=V_c+V_w\\\\M/\rho=M_c/\rho_c+Mw/\rho_w\\\\M/\rho=0.55*M/\rho_c+0.45*M/\rho_w\\\\1/\rho=0.55/\rho_c +0.45/\rho_w\\\\0.55/\rho_c=1/\rho-0.45/\rho_w\\\\0.55/\rho_c=1/(0.64*\rho_w)-0.45/\rho_w=(1/\rho_w)*(\frac{1}{0.64}-\frac{0.45}{1}  )\\\\0.55/\rho_c=1.1125/\rho_w\\\\\rho_c=\frac{0.55}{1.1125}*\rho_w= 0.494*\rho_w

The specific gravity of the wood chips is 0.494.

The average volume of a log is

V_l=(\pi*D^{2} /4)*L=(3.1416*\frac{8^{2}  \, in^{2} }{4} )*9ft*(\frac{12 in}{1ft})= 21714 in^{3}=12.57 ft^{3}

The weight of one log is

M=\rho*V=0.494*\rho_w*12.57  ft^{3}\\\\M=0.494*62.4\frac{lbm}{ft^{3} }*12.57ft^{3}\\\\M=387.5lbm

To provide 3617 ton/day of wood chips, we need

n=\frac{supply}{M_{log}}=\frac{3617 tons/day}{387.5 lbm}*\frac{2204lbm}{1ton}\\\\n=20573 logs/day=14.3 logs/min

The feed rate of logs is 14.3 logs/min.

7 0
2 years ago
Read 2 more answers
If you weigh 100 kg, how much would you weigh if all the water were removed from your body? A65 kg B45 kg C50 kg D35 kg
Mekhanik [1.2K]

Answer:

B. 45k

The human body is about 60 to 70% water.

(:

5 0
2 years ago
Write the net ionic equation for the reaction of zinc metal with aqueous iron(II) nitrate. Include physical states.
kotykmax [81]

Answer:

Zn°(s) + Fe⁺²(aq)  => Zn⁺²(aq) + Fe°(s)

Explanation:

Molecular Equation:

Zn°(s) + Fe(NO₃)₂(aq) => Zn(NO₃)₂(aq) + Fe°(s)

Ionic Equation:

Zn°(s) + Fe⁺²(aq) + 2NO₃⁻(aq) => Zn⁺²(aq) + 2NO₃⁻(aq) + Fe°(s)

Net Ionic Equation: => Drop NO₃⁻ as spectator ion

Zn°(s) + Fe⁺²(aq)  => Zn⁺²(aq) + Fe°(s)

7 0
2 years ago
Acrylonitrile () is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. If 1
pychu [463]

2C3H6 (g) + 2NH3 (g) + 3O2 (G) -> 2C3H3N (g) + 6H2O (g)

First off.. not a chem board.. but n e way.

This is a limiting reagent problem.

set it up as a DA problem.(Dimension Analysis)

Start with what you want.

you want Grams of acrylonitrile (C3H3N)

so start with that (Using ACL in place of Acrylonitrile.. just for ease of typing)

(g) = (53 g of ACL/1mol ACL) (2 mols ACL/2 mol C3H6)/ (1mol C3H6/42 grams) (15.0 grams)

solve that you wiill get grams of Acrylonitrile created by 15 grams oc C3H6 = 18.9g

Same setup for the two other reactants.

so i did it and for

oxygen I got 11.04 grams

and for Ammonia i got 15.29 grams

So the most you can make is 11.04 grams because if you have ot make any more .. you will have to get more O2 .. but since you have only 10 grams of it .. that is the most u can make in this reaction.

Both the other reactants are in excess.

rate brainliest pls

3 0
2 years ago
Isopentyl acetate (C7H14O2), the compound responsible for the scent of bananas, can be produced commercially. Interestingly, bee
lora16 [44]

Answer:

The answer to your question is:

a) 4.64 x 10 ¹⁵ molecules

b) 9.28 x 10 ¹⁵ atoms of O2

Explanation:

MW C7H14O2 = 84 + 14 + 32 = 130 g

a)        130 g of C7H14O2 ---------------- 1 mol of C7H14O2

           1 x 10 ⁻⁶ g              ---------------      x

           x = 7.7 x 10 ⁻⁹ mol

          1 mol of C7H14O2   --------------   6 .023 x 10 ²³ molecules

          7.7 x 10⁻⁹ mol          --------------    x

          x = 4.64 x 10¹⁵ molecules

b)      130 g of C7H14O2   ----------------   1 mol of C7H14O2

         1 x 10⁻⁶  C7H14O2   -----------------     x

         x = 7.7 x 10 ⁻⁹ mol of C7H14O2

        1 mol of C7H14O2    ---------------   2 mol of O2

        7.7 x 10 ⁻⁹                 ----------------   x

         x = 1.54 x 10⁻⁸ mol of O2

       1 mol of O2 -----------------  6.023 x 10 ²³ atoms

       1.54 x 10 ⁻⁸  ----------------   x

        x = 9.28 x 10 ¹⁵ atoms of O2

8 0
2 years ago
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