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Arlecino [84]
3 years ago
15

The density of copper is listed as 8.94g/cm​ 3​ . Two students each make three density determinations of samples of the substanc

e. Student A's results are 7.3g/mL, 9.4g/mL, and 8.3g/mL. Student B's results are 8.4g/cm​ 3​ , 8.8g/cm​ 3​ , 8.0g/cm​ 3​ . Compare the two sets of results in terms of precision and accuracy.
Chemistry
1 answer:
amm18123 years ago
5 0

Answer:

Density of the copper = 8.94g/cm^3

Student A results = 7.3gm/cm^3 ,9.4 gm/cm^3 , 8.3gm/cm^3

Student B results = 8.4 gm/cm^3 , 8.8 gm/cm^3 , 8gm/cm^3

From the observations we conclude that

Student A's result is accurate but not  precise as the trials noted are not close to each other.

Student B's result is accurate and precise as the trials noted are close to each other.

Mean density of student A = 7.3 + 9.4 + 8.3 /3 = 8.33gm/cm^3

Mean density of student B = 8.4 + 8.8 + 8 /3 = 8.4 gm/cm^3

both the densities of A and B are 0.5 away from the actual density.

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For the reaction a2 + b2 → 2ab, ea(fwd) = 105 kj/mol and ea(rev) = 77 kj/mol. assuming the reaction occurs in one step, calculat
Lady bird [3.3K]
Answer is: 28 kJ.
Chemical reaction: A₂ + B₂ ⇄ 2AB.
Ea(forward) = 105 kJ/mol.
Ea(reverse) = 77 kJ/mol.
ΔH(reaction) = ?
<span>The enthalpy change of reaction is the change in the energy of the reactants to the products.
</span>ΔH(reaction) = Ea(forward) - Ea(reverse).
ΔH(reaction) = 105 kJ/mol - 77 kJ/mol.
ΔH(reaction) = 28 kJ/mol; this is endothermic reaction (ΔH <span>> 0).</span>

6 0
3 years ago
A piece of high-density Styrofoam measuring 24.0 cm by 36.0 cm by 5.0 cm floats when placed in a tub of water. When a 1.5 kg boo
zheka24 [161]

Explanation:

The given data is as follows.

         Width of Styrofoam = 24.0 cm

          Length of Styrofoam = 36.0 cm

          Height of Styrofoam = 5.0 cm

Therefore, volume of the Styrofoam will be calculated as follows.

                  Volume = length × width × height

                                =  (36.0 × 24.0 × 5.0) cm^{3}

                                 = 4320 cm^{3}

or,                             = 4.32 \times 10^{3} cm^{3}

As Styrofoam partially sinks at 3.0 cm and total height of Styrofoam is 5.0 cm. Hence, height of Styrofoam above the water is (5.0 - 3 cm) = 2 cm.

So, volume of water displaced is as follows.

          24.0 cm × 36.0 cm × 2.0 cm

         = 1.73 \times 10^{3} cm^{3}

Hence, mass of displaced water is as follows.

                 mass = density × volume

                           = 1.00 g/cm^{3} \times 1.73 \times 10^{3} cm^{3}

                           = 1.73 \times 10^{3} g

Since, book is placed on the Styrofoam. Therefore, mass of water displaced is also equal to the following.

             Mass of water displaced = mass of book + mass of Styrofoam

                  1.73 \times 10^{3} g = 1500 g + mass of Styrofoam

                   (1730 - 1500) g = mass of Styrofoam

                   mass of Styrofoam = 230 g

Therefore, calculate the density of Styrofoam as follows.

                   Density = \frac{mass}{volume}  

                                 = \frac{230}{4.32 \times 10^{3} cm^{3}}

                                 = 53.24 \times 10^{-3} g cm^{-3}

Thus, we can conclude that the density of Styrofoam is 53.24 \times 10^{-3} g cm^{-3}.

4 0
3 years ago
Fermeroben heres the messages COUGH
lord [1]

I completely understand your gripe. I also believe that you are the better option. It will only take time until she is yours, despite your parent's views. Keep it up

6 0
3 years ago
El agua del mar contiene aproximadamente un 3,0 % m/v de sal (NaCl, 58,44 g/mol), (asuma que es la única fuente de cloruros) si
Alchen [17]

Answer:

s = 4.41 g/L.

Explanation:

¡Hola!

En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:

PbCl_2(s)\rightleftharpoons 2Cl^-(aq)+Pb^{2+}(aq)

Lo cual hace que la expresión de equilibrio se calcule como:

Ksp=[Pb^{2+}][Cl^-]^2

Y que en términos de la solubilidad molar, s, se resuelve como:

1.6x10^{-5}=s(2s)^2\\\\1.6x10^{-5}=4s^3\\\\s=\sqrt[3]{\frac{1.6x10^{-5}}{4} } \\\\s=0.0159molPbCl_2/L

Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):

s=0.0159molPbCl_2/L*\frac{278.1gmolPbCl_2}{1molmolPbCl_2} \\\\s=4.41g/L

¡Saludos!

7 0
3 years ago
What can student do to increase the force of attraction of magnet on the iron(metal) block?
Fantom [35]
Rub the magnet on the iron an it will cause a stronger force of attraction. Hope this helped!
3 0
3 years ago
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