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Goshia [24]
3 years ago
6

B. How many molecules are in 25.0 grams of H2SO4?

Chemistry
1 answer:
Firdavs [7]3 years ago
4 0

Answer:

4.23 x 10^-25 molecules

Explanation:

25.0 g H2SO (\frac{1 mol H2SO4}{98.1 g H2SO4}) (\frac{6.023x10^{23}  molecules}{1 mol H2SO4} ) = 4.23 x10^-25 molecules

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A 15.0 g sample of nickel metal is heated to 100.0 degrees C and dropped into 55.0 g of water, initially at 23.0 degrees C. Assu
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Answer: The final temperature of nickel and water is  25.2^{o}C.

Explanation:

The given data is as follows.

   Mass of water, m = 55.0 g,

  Initial temp, (t_{i}) = 23^{o}C,      

  Final temp, (t_{f}) = ?,

  Specific heat of water = 4.184 J/g^{o}C,      

Now, we will calculate the heat energy as follows.

           q = mS \Delta t

              = 55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)

Also,

    mass of Ni, m = 15.0 g,

   Initial temperature, t_{i} = 100^{o}C,

   Final temperature, t_{f} = ?

 Specific heat of nickel = 0.444 J/g^{o}C

Hence, we will calculate the heat energy as follows.

          q = mS \Delta t

             = 15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)      

Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.

              q_{water}(gain) = -q_{alloy}(lost)

55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C) = -(15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C))

       t_{f} = \frac{25.9^{o}C}{1.029}

                 = 25.2^{o}C

Thus, we can conclude that the final temperature of nickel and water is  25.2^{o}C.

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2 years ago
An alpha particle is equivalent to the nucleus of an atom of which element?(1 point)
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Answer:

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Explanation:

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