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3 years ago

Alcohol and water have different melting points. which method would you use to separate them?

1 answer:

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The process of separating alcohol from water can be done in several different ways. The most familiar method is through heating the blended liquid. Since alcohol has a lower boiling temperature than water, it will rapidly become steam. It can then be condensed into a separate container. You can also freeze the alcoholic mixture, which allows for partial removal of the nonalcoholic components; what remains will be more rich in alcohol. Use ordinary table salt to separate isopropyl alcohol from water. The result will be a condensed isopropyl alcohol, not a drinking alcohol.

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L avanzar hacia la derecha por el período 5, el tamaño atómico, la energía de ionización y los electrones de valencia: A. dismin

rusak2 [61]

Answer:

Al avanzar hacia la derecha por el período 5, el tamaño atómico, la energía de ionización y los electrones de valencia: A. disminuye, aumenta y aumentan, respectivamente.

Explanation:

El radio atómico representa la distancia que existe entre el núcleo y la capa de valencia, es decir la más externa. Por medio del radio atómico es posible determinar el tamaño del átomo. En un período el tamaño atómico disminuye de izquierda a derecha pues en este sentido aumenta el número atómico aumentando la carga nuclear mientras que el efecto pantalla y el número de niveles permanecen constantes. En otras palabras, disminuye de izquierda a derecha debido a la atracción que ejerce el núcleo sobre los electrones de los orbitales más externos, disminuyendo así la distancia núcleo-electrón.

Al avanzar hacia la derecha por el período 5, el tamaño atómico disminuye.

La energía de ionización es la necesaria para remover un electrón a un átomo en estado gaseoso. Mientras más lejos del núcleo esté el electrón, es más fácil removerlo porque se necesita menos energía. Al aumentar el número atómico de los elementos de un mismo período, se incrementa la atracción nuclear sobre el electrón más externo, ya que disminuye el radio atómico y aumenta la carga nuclear efectiva sobre él. Entonces en un período, al aumentar el número atómico, la energía de ionización aumenta.

Entonces, al avanzar hacia la derecha por el período 5, la energía de ionización y los electrones de valencia aumenta.

Los electrones de valencia son los electrones que están en la última capa electrónica (llamados orbitales de valencia) y tienen una alta posibilidad de participar en una reacción química.

En cada período aparecen los elementos cuyo último nivel de su configuración electrónica coincide con el número del período, ordenados por orden creciente de número atómico. Por ejemplo, el período 3 incluye los elementos cuyos electrones más externos están en el nivel 3.

Los electrones de valencia aumentan en número a medida que se avanza en un período.

Entonces, al avanzar hacia la derecha por el período 5, los electrones de valencia aumentan.

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3 years ago

When humans burn fossil fuels, most of the carbon quickly enters the_______

Len [333]

Answer:

Atmosphere.

Explanation:

Carbon moves from fossil fuels to the atmosphere when fuels are burned. When humans burn fossil fuels to power factories, power plants, cars and trucks, most of the carbon quickly enters the atmosphere as carbon dioxide gas.

8 0

3 years ago

Read 2 more answers

A student titrated 25.0 cm3 portions of dilute sulfuric acid with a 0.105 mol/dm3 sodium hydroxide solution.The equation for the

Rzqust [24]

Answer:

This question is incomplete

Explanation:

This question is incomplete as the volume of the base that was used during the titration was not provided. However, the completed question is in the attachment below.

The formula to be used here is CₐVₐ/CbVb = nₐ/nb

where Cₐ is the concentration of the acid = unknown

Vₐ is the volume of the acid used = 25 cm³ (as seen in the question)

Cb is the concentration of the base = 0.105 mol/dm³ (as seen in the question)

Vb is the volume of the base = 22.13 cm³ (22.1 + 22.15 + 22.15/3)

nₐ is the number of moles of acid = 1 (from the chemical equation)

nb is the number of moles of base = 2 (from the chemical equation)

Note that the Vb was based on the concordant results (values within the range of 0.1 cm³ of each other on the table) of the student

Cₐ x 25/0.105 x 22.13 = 1/2

Cₐ x 25 x 2 = 0.105 x 22.13 x 1

Cₐ x 50 = 0.105 x 22.13

Cₐ = 0.105 x 22.13/50

Cₐ = 0.047 mol/dm³

The concentration of the sulfuric acid is 0.047 mol/dm³

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7 0

2 years ago

A drought hits the habitat of a semi-aquatic bird population. All ponds dry up, and fish populations decline. There are two grou

alexandr1967 [171]

Answer:

The population of the long-legged birds decreases.

Explanation:

The population of the short-legged birds increases whereas the population of long-legged birds decreases due to availability of food in that environment. The long-legged birds feed on fish whose population decreases due to drought conditions so the population of long-legged birds also decreases while on the other hand, the population of short-legged birds increases or remain the same because they feed on the insects and the insects are available in large amount and less affected by the drought conditions.

4 0

3 years ago

A second- order reaction of the type A + B -->P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.0

andriy [413]

Answer:

a) 16.2 dm^3/mol*h

b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)

Explanation:

We can use the integrated rate equation in order to obtain k.

For the reaction A+ B --> P the reaction rate is written as

Rate = $-\frac{dC_A}{dt} = -\frac{dC_B}{dt} = \frac{dC_P}{dt} = kC_AC_B$

If $C_{A0}$ and $C_{B0}$ are the inital concentrations and x the concentration reacted at time t, so $C_A=C_{A0} -x$ and $C_B=C_{B0} -x$ and the rate at time t is written as:

$-\frac{dx}{dt} =-k(C_{A0} -x)(C_{B0}-x)$

Separating variables and integrating

$\int\limits^x_0 {\frac{1}{(C_{A0}-x)(C_{B0}-x)} } \, dx = \int\limits^t_0 {k} \, dt$

The integral in left side is solved by partial fractions, it can be used integral tables

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}}{C_{A0}-x}-ln\frac{C_{B0}}{C_{B0}-x}) =kt$

Using logarithm properties (ln x - ln y = ln(x/y))

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a $C_{B0}$ of 0.05 mol dm^-3.

Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.

So, the values given are t= 1, $C_{A0}=0.075$ , $C_{B0}=0.05$ , $C_{B}=0.02$ , it implies that the quantity reacted, x, is 0.03 and $C_{A}=0.075-x = 0.045$ . Then, the value of k would be

$kt = \frac{1}{0.05-0.075}(ln\frac{0.075*0.02}{0.045*0.05})$

$k = 16.21 \frac{dm^3}{mol*1h}$

b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.

For reactant A, It is solved with the same equation

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

but suppossing that $C_A= C_{A0}/2=0.0375$ so $C_B=C_{B0}- C_{A0}/2=0.0125$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.0125}{0.0375*0.05})$

$t=1.71 h = 1.71*3600 s = 6.1*10^3 s$

For reactant B, $C_B= C_{B0}/2=0.025$ so $C_A=C_{A0}- C_{B0}/2=0.05$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.025}{0.05*0.05})$

$t=0.71 h = 0.7*3600 s = 2.5*10^3 s$

Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)

3 0

2 years ago