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Vinil7 [7]
3 years ago
13

a sample of liquid copper required 254 kj to vaporize the sample. Calculate the mass of copper vaporized. The heat of vaporizati

on of copper is 4.81 kj/g
Chemistry
1 answer:
NNADVOKAT [17]3 years ago
5 0

Answer:

Mass of copper vaporized is 52,8g

Explanation:

The heat of vaporization (ΔHvap) is defined as the amount of energy you require to transform a quantity of liquid in gas. The formula is:

Q = ΔHvap×mass

Where Q is the amount of heat and mass is the mass of the compound.

If you required 254 kJ to vaporize a sample of copper which heat of vaporization is 4,81 kJ/g:

254kJ = 4,81 kJ/g × mass of copper

<em>52,8g = mass of copper</em>

<em></em>

I hope it helps!

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If a mixture looks smooth and the same throughout it is probably what
gulaghasi [49]
Answer:
             <span>If a mixture looks smooth and the same throughout it is probably <u>Homogeneous</u>.

Explanation:
                   Mixture is the combination of different compounds which are unreactive to each other.

Mixture are classified as ...

Solutions; in which the mixed compounds are thoroughly mixed and cannot be distinguished from each other and are said to be homogeneous. In solutions the size of solute is very small (i.e. Less than 1 nm).

Colloids; in which the solute is homogeneous visually but heterogeneous microscopically. The size of particles in this case is between 1 nm to 1 </span>μm.

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4 0
3 years ago
Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a tank
Masja [62]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration equilibrium constant is K_c  = 14.39

Explanation:

The chemical equation for this decomposition of ammonia is

                2 NH_3  ↔   N_2 + 3 H_2

The initial concentration of ammonia is mathematically represented a

          [NH_3] =  \frac{n_1}{V_1}  = \frac{29}{75}

          [NH_3] = 0.387  \  M

The initial concentration of nitrogen gas  is mathematically represented a

         [N_2] =  \frac{n_2}{V_2}

         [N_2] =  0.173  \  M

So  looking at the equation

   Initially (Before reaction)

      NH_3 = 0.387 \ M

      N_2  =  0 \  M

      H_2 =  0 \ M

During reaction(this is gotten from the reaction equation )

        NH_3 = -2 x(this implies that it losses two moles of concentration )

         N_2 = + x  (this implies that it gains 1 moles)

         H_2  =  +3 x(this implies that it gains 3 moles)

Note : x denotes concentration

At equilibrium

        NH_3 = 0.387 -2x

       N_2 =  x

        H_2  =  3 x

Now since

     [NH_3] = 0.387  \  M

     x= 0.387  \  M    

H_2  =  3 * 0.173    

H_2  =  0.519 \ M    

NH_3 = 0.387 -2(0.173)

NH_3 = 0.041 \ M

Now the equilibrium constant is

           K_c  =  \frac{[N_2][H_2]^3}{[NH_3]^2}

substituting values

           K_c  =  \frac{(0.173) (0.519)^3}{(0.041)^2}

           K_c  = 14.39

         

3 0
3 years ago
Who was given name black hole​
marin [14]

Answer:

John Archibald Wheeler

3 0
3 years ago
The coolest stars in the sky are ____ in color
Mekhanik [1.2K]
Red is the answer in the blank
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3 years ago
Read 2 more answers
Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2
Setler79 [48]

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

7 0
3 years ago
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