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WITCHER [35]
3 years ago
6

-300g de acido comercial se disuelve en agua destilada contenidos en un cono, cuyo radio es de 0.005Km y 300cm de altura, si la

densidades de 1.2g/m3 ¿Cuál es la concentración expresada en %m/m?
Chemistry
1 answer:
Bond [772]3 years ago
8 0

Answer:

\%m/m=76.1\%

Explanation:

¡Hola!

En este caso, considerando la información dada, entendemos que se haría primero necesario calcular el volumen del cono en metros cúbicos, teniendo en cuenta que 0.005 km son 5 m y 300 cm son 3 m:

V=\frac{1}{3} \pi r^2h\\\\V=\frac{1}{3} \pi *(5m)^2(3m)=78.5m^3\\\\

Ahora, convertimos esta cantidad a gramos por medio de la densidad para conocer la masa de la solución:

m_{sol}=78.5m^3*\frac{1.2g}{1m^3} =94.2g

Finalmente, aplicamos la definición de %m/m para obtener:

\%m/m=\frac{300g}{300g+94.2g}*100\%\\\\ \%m/m=76.1\%

¡Saludos!

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Chlorine has two natural isotopes, chlorine-35 and chlorine-37.
Alex Ar [27]

Answer:

The relative atomic mass of chlorine depends on the ratio between the abundance of these two naturally-occurring isotopes.

Explanation:

The relative atomic mass of an element is a weighted average of the atomic mass of its naturally-occurring isotopes. The relative abundance of each isotope gives its weight in this weighted average.

For these two naturally-occurring isotopes of chlorine:

  • The relative atomic mass of ^{35}{\rm Cl} is approximately 34.969 daltons. The relative abundance of this isotope in nature is approximately 0.758.
  • The relative atomic mass ^{37}{\rm Cl} is approximately 36.966 daltons. The relative abundance of this isotope in nature is approximately 0.242.

\begin{array}{|c|c|c|}\cline{1-3} \text{Isotope} & \text{Atomic Mass} & \text{Relative Abundance}\\ \cline{1-3} ^{35}{\rm Cl} & \approx 34.968\; \rm Da} & \approx 0.758 \\ \cline{1-3} ^{37}{\rm Cl} & \approx 36.966\; \rm Da & \approx 0.242 \\ \cline{1-3}\end{array}.

\begin{aligned}&\text{relative atomic mass of Cl} \\ &= \text{atomic mass of $^{35}{\rm Cl}$} \times \text{relative abundance of $^{35}{\rm Cl}$} \\&\quad + \text{atomic mass of $^{37}{\rm Cl}$} \times \text{relative abundance of $^{37}{\rm Cl}$} \\ &\approx 34.968\; \rm Da \times 0.758 + 36.966\; \rm Da \times 0.242 \\ &\approx 35.45\; \rm Da \end{aligned}.

The relative abundance of ^{35}{\rm Cl} is much higher than that of ^{37}{\rm Cl}. Consequently, the relative atomic mass of the element \rm Cl is closer to the atomic mass of ^{35}{\rm Cl}\! than that of ^{37}{\rm Cl}\!.

7 0
3 years ago
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natita [175]

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3 0
3 years ago
4. Which two compounds contain the same total number of atoms?
Alja [10]

Answer:

b

Explanation:

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3 years ago
If the NaOH is added to 35.0 mL of 0.167 M Cu(NO3)2 and the precipitate isolated by filtration, what is the theoretical yield of
Ghella [55]

Answer:

The correct answer is - 0.570 grams

Explanation:

The balanced chemical reaction is given by

Cu(NO3)2(aq)     + 2NaOH(aq)    -------->    Cu(OH)2(s)      + 2NaNO3(aq)

        1.0 mole            2.0 mole                 1.0 mole          2.0 mole

number of mol of Cu(OH)2,

n = Molarity * Volume

= 35.0*0.167 = 5.845 millimoles

As clear in the equation, 1 mole of Cu(NO3)2 gives 1 mole of Cu(OH)2 , So, 5.845 millimoles of Cu(NO3)2 will produce 5.845 millimoles of Cu(OH)2

Mass of Cu(OH)2 = number of mol * molar mass

= 97.5*5.845*10^-3

= 0.570 grams

Thus, the correct answer is - 0.570 grams

8 0
3 years ago
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