Which of the following does not describe a characteristic of carbon

What is it called when the moon blocks the view from the sun?

defon

Answer:

solar eclipse

Explanation:

8 0

3 years ago

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(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g

Anarel [89]

Answer:

Part A

The volume of the gaseous product is $V = 787L$

Part B

The volume of the the engine’s gaseous exhaust is $V_e = 2178 \ L$

Explanation:

Part A

From the question we are told that

The temperature is $T = 350^oC = 350 +273 =623K$

The pressure is $P = 735 \ torr = \frac{735}{760} = 0.967\ atm$

The of $C_8 H_{18} = 100.0g$

The chemical equation for this combustion is

$2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}$

The number of moles of $C_8 H_{18}$ that reacted is mathematically represented as

$n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }$

The molar mass of $C_8 H_{18}$ is constant value which is

$M = 114.23 \ g/mole$

So $n = \frac{100 }{114.23} }$

$n = 0.8754 \ moles$

The gaseous product in the reaction is $CO_2_{(g)}$ and water vapour

Now from the reaction

2 moles of $C_8 H_{18}$ will react with 25 moles of $O_2$ to give (16 + 18) moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So

1 mole of $C_8 H_{18}$ will react with 12.5 moles of $O_2$ to give 17 moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

This implies that

0.8754 moles of $C_8 H_{18}$ will react with (12.5 * 0.8754 ) moles of $O_2$ to give (17 * 0.8754) of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So the no of moles of gaseous product is

$N_g = 17 * 0.8754$

$N_g = 14.88 \ moles$

From the ideal gas law

$PV = N_gRT$

making V the subject

$V = \frac{N_gRT}{P}$

Where R is the gas constant with a value $R = 0.08206 \ L\cdot atm /K \cdot mole$

Substituting values

$V = \frac{14.88* 0.08206 *623}{0.967}$

$V = 787L$

Part B

From the reaction the number of moles of oxygen that reacted is

$N_o = 0.8754 * 12.5$

$N_o = 10.94 \ moles$

The volume is

$V_o = \frac{10.94 * 0.08206 *623}{0.967}$

$V_o = 579 \ L$

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

$V_e = V_o * \frac{0.79}{0.21}$

Substituting values

$V_e = 579 * \frac{0.79}{0.21}$

$V_e = 2178 \ L$

3 0

3 years ago

Where is the atmosephere the most dence

likoan [24]

The Earths Surface, it’s also known as the troposphere also dense* lol

8 0

3 years ago

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(c) Polar solutes were separated by hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase. How would

Lelu [443]

Answer:

(c) The retention time would be higher (d) The retention time would be lower.

Explanation:

For the polar solutes which were separated using the hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase, the retention time would be higher if eluent were changed from 80 vol% to 90 vol% acetonitrile in water.

However, for the polar solutes which were separated using the normal-phase chromatography on bare silica with methyl t=butyl ether and 2-propanol solvent, the retention time would be lower if the eluent were changed from 40 vol% to 60 vol% 2-propanol.

5 0

3 years ago

A space air is at a temperature of 75 oF, and the relative humidity (RH) is 45%. Using calculations, find: (a) the partial press

earnstyle [38]

Answer:

A) Partial Pressure of dry air = 13.32 KPa

Partial Pressure of water vapour = 1.332 KPa

B) Humidity ratio; X = 0.0691

C) V_p = 0.8384 m³/Kg

Explanation:

A) We are given;

Temperature = 75°F

Relative Humidity = 45%

Now,to calculate the partial pressure, we will use the relationship;

Relative Humidity = (Partial Pressure/Vapour Pressure) × 100%

Making partial pressure the subject;

Partial Pressure = Relative Humidity × Vapour Pressure/100%

From the first table attached, at temperature of 75°F, the vapor pressure is 29.6 × 10^(-3) bar = 29.6 KPa

Thus;

Partial Pressure of dry air = (45 × 29.6)/100

Partial Pressure of dry air = 13.32 KPa

From online values, vapour pressure of water vapour at 75°F = 2.96 KPa

Thus;

Partial Pressure of water vapour = (45 × 2.96)/100 = 1.332 KPa

B) humidity ratio of moist air is given as;

X = 0.62198 pw / (pa - pw)

where;

pw = partial pressure of the water vapor in moist air

pa = atmospheric pressure of the moist air

Thus;

X = (0.62198 × 1.332)/(13.32 - 1.332)

X = 0.0691

C) Formula for moist air specific volume is;

V_p = (1 + (xRw/Ra) × RaT/p

Where;

V_p is specific volume

T is temperature = 75°F = 297.039 K

p is barometric pressure which in this case is standard sea level pressure = 101.325 KPa

pw is partial pressure of the water vapor in moist air = 1.332 KPa

Rw is individual gas constant for water = 0.4614 KJ/Kg.K

Ra is individual gas constant for air = 0.2869 KJ/Kg.K

V_p = (1 + (0.0691 * 0.4614/0.2869)) × 0.286.9 * 297.039/101.325

V_p = 0.8384 m³/Kg

6 0

3 years ago