. If 69.15% exists as 63Cu with an atomic mass of 62.92, then what is the atomic mass of the second isotope?

What two types of energy does a waterfall have

Mrac [35]

Kinetic Energy which relies on an objects mass and velocity and Potential Energy which relies on the height of the object

3 0

4 years ago

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Which of the following statements is true?

eduard

Answer:

Lines of latitude are used to divide Earth into 24 time zones

Explanation:

5 0

3 years ago

A balloon is filled with 6.5 L of helium gas at 25oC and 1.0 atm. What volume will the balloon have when the temperature is lowe

harkovskaia [24]

Figure it out yet ?????

5 0

3 years ago

1. Take the reaction: NH3 + O2 + NO + H2O. In an experiment, 3.25g of NH3 are allowed

Rudiy27

Answer:

5.74g of NO

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

4NH3 + 5O2 —> 4NO + 6H2O

Step 2:

Determination of the masses of NH3 and O2 that reacted and the mass of NO produced from the balanced equation. This is illustrated below:

Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g

Molar Mass of NO = 14 + 16 = 30g/mol

Mass of NO from the balanced equation = 4 x 30 = 120g

From the balanced equation above,

68g of NH3 reacted with 160g of O2 to produce 120g of NO.

Step 3:

Determination of the limiting reactant.

We need to determine the limiting because it will be used to calculate the maximum yield of the reaction. This is illustrated below:

From the balanced equation above,

68g of NH3 reacted with 160g of O2.

Therefore, 3.25g of NH3 will react with = (3.25 x 160)/68 = 7.65g of O2.

From the simple illustration above, we can see that lesser mass of O2 is needed to react with 3.25g of NH3. Therefore, NH3 is the limiting reactant while O2 is the excess reactant.

Step 4:

Determination of the mass of NO produced from the reaction.

In this case the limiting reactant will be used because all of it were used in the reaction.

The limiting reactant is NH3.

From the balanced equation above,

68g of NH3 reacted to produce 120g of NO.

Therefore, 3.25g of NH3 will react to produce = (3.25 x 120)/68 = 5.74g of NO.

From the calculations made above, 5.74g of NO is produced.

4 0

3 years ago

El pH de una solución acuosa 10^-14 M de ácido acético, a 25°C, es igual a, teniendo en cuenta que Ke = 1.76x10^-5

frez [133]

Answer:

$pH=14$

Explanation:

Hola!

En este caso, consideramos que la disociación de ácido acético ocurre:

$CH_3COOH\rightarrow CH_3COO^-+H^+$

Así, mediante la solución del equilibrio ácido, podemos calcular la concentración de iones hidronio que posteriormente sirven para calcular el pH de la solución, por tal razón, debemos calcular el equilibrio dada la constante de equilibrio y por medio de la ley de acción de masas en términos del cambio $x$ como cualquier problema de equilibrio:

$Ke=\frac{CH_3COO^-][H^+]}{[CH_3COOH]}\\\\1.76x10^{-5}=\frac{x*x}{1x10^{-14}M-x}$

Resolviendo para $x$ , tenemos $x=0.999x10^{-14}$

Así, la concentración de hidrógeno es igual a x, por lo que el pH:

$pH=-log([H^+])=-log(0.999x10^{-14})\\\\pH=14$

Dicho valor tiene sentido desde que la concentración de hidrógeno es casi despreciable, por lo que se puede asumir que tiende a ser básica.

Saludos!

5 0

3 years ago

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