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2 years ago

What is the mass of a 32.5-ml sample of ethanol? the density of ethanol is 0.789 g/ml.

Chemistry

1 answer:

Anon25 [30]2 years ago

6 0

Answer:

25.6 grams of ethanol in 32.5 ml of the substance.

Explanation:

The density of ethanol can be used as a conversion factor. We know that:

Ethanol is 0.789 g/ml. That becomes a conversion factor that we may write as (0.789 g/ml) and then convert either grams ethaanol or ml ethanol into the other unit. We are given volume of ethanol: 32.5 ml.

Note that when we multiply (0.789 g/ml) by (32.5 ml) the ml unit cancels, leaving just grams. That's what we want, so:

(0.789 g/ml)(32.5 ml) = 25.6 grams of ethanol in 32.5 ml of the substance.

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Why gold is called inert metal

astraxan [27]

Gold has a heavy enough nucleus that its electrons must travel at speeds nearing the speed of light to prevent them from falling into the nucleus. This relativistic effect applies to those orbitals that have appreciable density at the nucleus, such as s and p orbitals. These relativistic electrons gain mass and as a consequence, their orbits contract. As these s and (to some degree) p orbits are contracted, the other electrons in d and f orbitals are better screened from the nucleus and their orbitals actually expand.

Since the 6s orbital with one electron is contracted, this electron is more tightly bound to the nucleus and less available for bonding with other atoms. The 4f and 5d orbitals expand, but can't be involved in bond formation since they are completely filled. This is why gold is relatively unreactive.

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5 0

3 years ago

Unit conversions.(a) Convert 10,000 dynes to units of lbm ∙ ft/s^2 and lbf(b) Convert 0.2 atm to units of Kpa and lbf/in^2(c) Co

Maru [420]

Answer:

(a) 0.72 lbm· ft/s²

(b) 20.3 kPa, 2.94 lbf / in²

(d) 1.5 x 10⁻² J, 6.1 x 10⁻² cal

Explanation:

Our strategy here will be to find the conversion factors for the quantities we are asked in each part, and perform the calculations.

(a) 10,000 dynes to lbm ·ft/s²

here we are asked to convert the force of 10,000 dynes to lbm ·ft/s². Recall that F= ma ( m= mass, a = acceleration), thus

10,000 dynes = 10 g cm/s²

converting the force

10,000 g cm/s² x (1 lbm/454 g) x (1 ft / 30.48 cm ) /s² = 0.72 lbm· ft/s²

(b)

1 atm = 101.33 pa

0.2 atm x ( 101.33 kPa ) = 20.3 kPa

1 atm = 14.7 lbf / in²

0.2 atm x ( 14.7 lbf / in² /atm ) = 2.94 lbf / in²

ºF = 9/5 ºC +32

ºF = 9/5 ( 37ºC) + 32 = 98.6 ºF

K = ºC + 273

K = (37 + 273) K = 310 K

(d) 50 in²·lbm/s² to joules and calories

Since the unit in² ·lbm/s² is not that common, lets convert it using their definition.

These are energy units, and we know the energy is the force times distance. In turn force is mass times acceleration so that the units of energy are mass time distance per time squared.

Joules is the unit of energy in the metric system.

50 in² lbm/s² = 50 in²x ( 2.54 cm/in x 1m /100cm)² x (1lbm x 0.454 Kg/lbm)/s²

= 1.5 x 10⁻² Kg m²/² = 1.5 x 10⁻² J

To convert to cal it wilñl be easier to use the value in joules just calculated:

1.5 x 10⁻² J x (4.184 cal/J) = 6.1 x 10⁻² cal

4 0

3 years ago

Balance the equation <br> NaOH + CH3COOH

solong [7]

NaOH + CH3COOH -> CH3COONa + H20

8 0

3 years ago

Problems that I am stuck on! Need the answers asap

Anna71 [15]

Answer:

1.b

2.b

3.a

4.a

Explanation:

use the rate laws when answering this.

6 0

2 years ago

A una mezcla de 300g, formada con 60% P/P de Hierro y 40% P/P de Arena, se le adicionan 135g de Cobre y 2,77g de Aluminio. ¿Cuál

Svetllana [295]

Answer:

$\%P/P_{hierro}=41.1\%\\\\\%P/P_{arena}=24.4\%\\\\\%P/P_{cobre}=30.8\%\\\\\%P/P_{aluminio}=0.6\%$

Explanation:

¡Hola!

En este caso, dado que estamos tratando con problem sobre porcentaje peso/peso de hierro, arena, cobre y aluminio, primero debemos calcular la masa inicial de estos dos primeros en la mezcla original de acuerdo con:

$m_{hierro}=300g*0.60=180g\\\\m_{arena}=300*0.40=120g$

Ahora si podemos calcular la masa de la mezcla final como la suma de las masas de todos los constituyentes de la mezcla:

$m_T=180g+120g+135g+2.77g=437.77g$

Finalmente, podemos calcular los porcentajes P/P como se muestra a continuación:

$\%P/P_{hierro}=\frac{180g}{437.77g} *100\%=41.1\%\\\\\%P/P_{arena}=\frac{120g}{437.77g} *100\%=24.4\%\\\\\%P/P_{cobre}=\frac{135g}{437.77g} *100\%=30.8\%\\\\\%P/P_{aluminio}=\frac{g}{437.77g} *100\%=0.6\%$

¡Saludos!

5 0

3 years ago