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Novay_Z [31]
3 years ago
5

In a combustion chamber, ethane (C2H6) is burned at a rate of 8 kg/h with air that enters the combustion chamber at a rate of 17

6 kg/h. Determine the percentage of excess air used during this process.
Chemistry
1 answer:
REY [17]3 years ago
4 0

Answer:

37%

Explanation:

From the question, the equation goes does.

C2H6+ (1-x)+a(O2+3.76N2)=bC02 + cH2O + axO2 + 3.76dN2.

Mair=Mair/Rin

( MN)O2 + (MN)N2÷ (MN)O2 + (MN)N2 +(MN)C2H6.

33 . 3.25(1-x) + 28 × 13.16(1-x) ÷ 33 × 3.25(1-x) + 28 × 13.16(1-x). + 30.1

= 176/176+8

X= 0.37

0.37 × 100

X= 37%

You might be interested in
Write the concentrated sulphuric acid H2C2O42H2O=2H2O+CO2+CO
masha68 [24]

Answer:

H2C2O4.2H20 → CO2 + CO + H2O

Explanation:

Oxalic acid crystals are nothing but dehydrated oxalic acid (H2C2O4 . 2H2O).

On heating, the water of crystallization is lost first. Then, the dehydrated oxalic acid decomposes into carbon dioxide(CO2), carbon monoxide(CO) and water(H2O).

Equations involved :

H2C2O4 . 2H2O → H2C2O4 + 2H2O

H2C2O4 → CO2 + HCOOH (FORMIC ACID)

HCOOH → CO + H2O

Overall equation : H2C2O4.2H20 → CO2 + CO + H2O

4 0
3 years ago
if the volume of a gas contracted from 648 mL to 0.15L, what was its final pressure if it started at a pressure of 485 kpa
bija089 [108]

Answer:

2100 kPa

Explanation:

The temperature is constant, so the only variables are pressure and volume.

We can use Boyle’s Law.

p₁V₁ = p₂V₂     Divide both sides of the equation by V₂

p₂ = p₁ × V₁/V₂  

p₁ = 485 kPa; V₁ =              648 mL  

p₂ = ?;            V₂ = 0.15 L = 150 mL      Calculate p₂

p₂ = 485 × 648/150

p₂ = 2100 kPa

4 0
3 years ago
How do i convert decimals into scientific notation
zloy xaker [14]

Answer:

To convert a decimal into scientific notation, move the decimal point until you get to the left of the first non-zero integer. The number of places the decimal point moves is the power of the exponent, because each movement represents a "power of 10".

5 0
3 years ago
A buffer contains significant amounts of ammonia and ammonium chloride. part a write an equation showing how this buffer neutral
siniylev [52]
Hello!

When HI is added, the buffer reacts in the following way:

1) Neutralizing of the Acid:

HI + NH₄OH → NH₄I + H₂O

2) Dissociation of the salt of a weak acid:

NH₄I → NH₄⁺(aq) + I⁻ (aq)

3) Dissociation of a weak acid to form H₃O⁺ (very little):

NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺

This series of reactions show how the adding of a strong Acid can be neutralized by the buffer, releasing instead very little amounts of Hydronium ions.

Have a nice day!
8 0
3 years ago
A gas with a volume of 2 L at 25°C is placed into a container that is 4 L. What is the new temperature of the gas?
zalisa [80]

Answer:

Convert 539 torr to atm.

2.) A gas takes up 25.2 liters at 25oC. At 25oC, the gas can also take up 12.2 liters at 1500 torr. What was the pressure, in atm, of the original sample?

3.) A gas takes up 14.8 liters of 24oC. What temperature in kelvin is required to obtain a volume of 25.0 liters at constant pressure?

4.) How many moles of chlorine gas are present at 25oC, 762 torr, with a volume of 14.2 L?

5.) If "Bor" (thats the name of a person) has a sample of gas that has a volume of 8.2 liter at 25oC and 2 atm, how much volume will it take up if you decrease pressure to 1.5 atms and icrease temperature to 100oC?

6.) 25 liters of gas A is pumped into a container at 25oC and 760 torr with 20 liters of gas B at 25oC and 700 torr. Calculate the total pressure when both gases are pumped into a tank with 10 liters at 25oC.

7.) Calculate the mole fraction of oxygen when 200 torr of air (760 torr total) is oxygen.

8.) KClO3 is decomposed by the following reaction:

2KClO3(s) ---> 2KCl(s) + 3O2(g)

The O2 produced was collected by the displacement of water at 22oC at a total pressure of 760 torr. The volume of gas collected was 1.20 liters, and the vapor pressure of water at 22oC is 21 torr. Calculate the partial pressure of O2 in the gas collected and the mass of KClO3 in the sample that was decomposed.  

9.) Calculate the root mean square velocity of hydrogen gas at 25oC.

10.) What is the ratio of effusion between fluorine and chlorine?

1.) Just set up a simple ratio.

539 torr * (1 atm/760 torr) = 0.709 atm

2.) Use this formula:

P1V1 = P2V2

460 atm  * 1.20 liters = 839 torr * V2

V2= 0.658 L

3.) You will use the ratio P1V1 = P2V2 in order to solve this equation.

P1V1 = P2V2

25.2 liters * P1 = 12.2 liters * (1500 torr / 760 torr)

Note that you have to convert the torr to atm. The problem asks for the final pressure in atms. You could of just solved the ratio without the conversion, and converted the answer later. Either way is acceptable.

25.2 liters * P1 = 24

P1 = 0.956 atm

4.) Use the formula:

V1/T1 = V2/T2

14.8 liters / 297 torr = 25.0 liters / x

14.8 x = 7425

x = 502 K

5.) PV = nRT (the ideal gas law! Note: you will see this a LOT in gases, so memorize it)

n = PV/RT

P = (762 torr/760 torr) = 1.00atm

n = (1.00 atm * 14.2 L) / (298oK * 0.08206)

n = 0.582 moles

6.) PV = nRT

n = PV/RT

= (2 atm * 8.2 L) / (0.08206 * 298oK) = 0.671 moles

Now, you must remember that you are taking this amount and changing it some more. So, you do this again, this time, solving for VOLUME. (No, the initial 8.2 doesn't count, because you are manipulating the environment)

PV = nRT

V = nRT/P

=(0.671 moles * 0.08206 * 373oK) / 1.5 atm = 13.76 liters

7.) You must solve this twice for both gas A and B, then combine the individual pressures to obtain the final pressure. Use the Ideal Gas Law.

ngas a = PV/RT = (1 atm * 25 liters) / (0.08206 * 298 K) = 1.02 moles

ngas b = PV/RT = (1 atm * 20 liters) / (0.08206 * 298) = 0.818 moles

You know how many moles you have. The second part of the problem calls for a manipulation of the environment, which means that you will be solving for PRESSURE.

Pgas a = nRT/V = (1.02 moles * 0.08206 * 298K / 10.0 liters) = 2.49 atm

Pgas b = nRT/V = (0.818 moles * 0.08206 * 298 K / 10.0 liters) = 2.00 atm

2.49 atm + 2.00 atm = 4.49 atm

8.) This is a simple mole fraction. You know that air has a total of 760 torr. Since we are given that O2 directly contributes 200 of that, we can set up this ratio:

O2 pressure/ total pressure = 200 torr / 760 torr = 0.263 = 26.3%

9.) Remember that:

Ptotal = PO2 + PH2O = PO2 + 21 torr = 760 torr

PO2 = 739 torr

PV = nRT or n=PV/RT

P = 739 torr/760 torr = 0.972 atm

n = (0.972 atm * 1.20 liters) / (0.08206 * 295 K) = 0.0482 moles O2

The final answer calls for mass of the KClO3, not the O2. So we must find the number of moles of KClO3.

0.0482 moles O2 * (2 mol KClO3/3 mol O2) = 0.0321 moles KClO3

0.0321 mol KClO3 * (122.6 g KClO3 / 1 mol KClO3) = 3.94 grams KClO3

10.) Remember that urms formula is:

So, using that, let's fill in our variables and constants. R, for this equation is 8.3145 J/K*mol and T = 298 K.  Calculating for M =

MH2 = 1 g/mol * 2 mol (since H2 is diatomic) * 1 kg/1000g = 0.002 kg

(3 * 8.3145 * 298/ 0.002) = 3716581.5 (you have to take the square root of this) -----> 1928 m/s

11.) Both fluorine and chlorine are diatomic. The equation relating effusion and diffusion is:

Taking this, you take the square root of 70.90 g (which is the weight of Cl2), and divide it by the square root of 37.996 g (which is the weight of F2). The answer is 1.366. This means that fluorine effuses 1.366 times as fast as chlorine, which makes sense because F2 is smaller.

Explanation:

3 0
3 years ago
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