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Ilya [14]
2 years ago
11

Bromine and nitrogen combine to form nitrogen tribromide (NBr3). How many liters of NBr3 gas are produced when 6.85 L of bromine

reacts with excess nitrogen? Assume conditions are at STP.
Chemistry
1 answer:
dlinn [17]2 years ago
7 0

Answer:

4.567 ltr

Explanation:

both are gases

so

N2 +Br2 =NBr3

balancing

N2 +3B2 = 2 NBr3

now

3 litr of B2 gives 2 liter of NBr3

6.85 litr of B2 gives 2/3 *6.85 litr NBr3 = 4.567ltr

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Which of the following combustion reactions is balanced correctly? A. C4H6 + 5.5O2 4CO2 + 3H2O B. C4H6 + 4O2 4CO2 + 3H2O C. C4H6
Airida [17]
You must verify that the number of atoms of each type is equal on both sides of the chemical equation: same number of C, same number of H and same number of O on both sides.

<span>A. C4H6 + 5.5O2 ---> 4CO2 + 3H2O

element      reactant side      product side

C                4                        4
H                6                        3*2 = 6
O                5.5 * 2 = 11        4*2 + 3 = 11

Then, this equation is balanced.

</span>Do the same with the other equations if you want to verify that they are not balanced.

Answer: option A.
5 0
3 years ago
Why does nobody fking answer my questions.
raketka [301]

Answer:

by wearing of rocks

Explanation:

earths gravity your welcome

6 0
2 years ago
Read 2 more answers
.an open flask contains 0.200 mol of air. atmospheric pressure is 745 mmhg and room temperature is 68˚f. how many moles are pres
aleksklad [387]

<u>0.219 moles </u><u>moles are present in the flask when the </u><u>pressure </u><u>is 1.10 atm and the temperature is 33˚c.</u>

What is ideal gas constant ?

  • The ideal gas constant is calculated to be 8.314J/K⋅ mol when the pressure is in kPa.
  • The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas.
  • The combined gas law relates pressure, volume, and temperature of a gas.

We simple use this formula-

The basic formula is PV = nRT where. P = Pressure in atmospheres (atm) V = Volume in Liters (L) n =  of moles (mol) R = the Ideal Gas Law Constant.

68F = 298.15K

V = nRT/P = 0.2 * 0.08206 * 298.15K / (745/760) = 4.992Liters

n = PV/RT = 1.1atm*4.992L/(0.08206Latm/molK * 306K)

n = 0.219 moles

Therefore, 0.219 moles moles are present in the flask when the pressure is 1.10 atm and the temperature is 33˚c.

Learn more about ideal gas constant

brainly.com/question/3961783

#SPJ4

4 0
1 year ago
Gaseous butane, CH3(CH2)2CH, reacts with gaseous oxygen gas, O2, to produce gaseous carbon dioxide, CO2, and gaseous water, H2O.
weeeeeb [17]

Answer:

Percentage yield of carbon dioxide is 49.9%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2CH3(CH2)2CH3 + 13O2 —> 8CO2 + 10H2O

OR

2C4H10 + 13O2 —> 8CO2 + 10H2O

Next, we shall determine the masses of butane and oxygen that reacted and the mass of carbon dioxide produced from the balanced equation. This is illustrated below:

Molar mass of butane C4H10 = (12×4) + (10×1)

= 48 + 10

= 58 g/mol

Mass of C4H10 from the balanced equation = 2 × 58 = 116 g

Molar mass of O2 = 16 × 2 = 32 g/mol

Mass of O2 from the balanced equation = 13 × 32 = 416 g

Molar mass of CO2 = 12 + (16×2)

= 12 + 32

= 44 g/mol

Mass of CO2 from the balanced equation = 8 × 44 = 352 g

Summary:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen to produce 352 g of carbon dioxide.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen.

Therefore, 34.29 g of butane will react with = (34.29 × 416) / 116 = 122.97 g of oxygen.

From the calculation made above, we can see clearly that only 122.97 g out of 165.7 g of oxygen reacted completely with 34.29 g of butane. Therefore, butane is the limiting reactant and oxygen is the excess reactant.

Next, we shall determine the theoretical yield of carbon dioxide.

In this case, we shall use the limiting reactant because it will give the maximum yield of carbon dioxide as all of it is used up in the reaction.

The limiting reactant is butane and the theoretical yield of carbon dioxide can be obtained as follow:

From the balanced equation above,

116 g of butane reacted to produce 352 g of carbon dioxide.

Therefore, 34.29 g of butane will react to produce = (34.29 × 352) / 116 = 104.05 g of carbon dioxide.

Therefore, the theoretical yield of carbon dioxide is 104.05 g

Finally, we shall determine the percentage yield of carbon dioxide as follow:

Actual yield of carbon dioxide = 51.9 g

Theoretical yield of carbon dioxide = 104.05 g

Percentage yield of carbon dioxide =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield of carbon dioxide = 51.9 / 104.05 × 100

Percentage yield of carbon dioxide = 49.9%

7 0
3 years ago
What are the examples of physical chemistry?
OverLord2011 [107]

ANSWER:

Physical chemistry is the branch of chemistry that deals with the physical structure of chemical compounds, the way they react with other matter and the bonds that hold their atoms together. An example of physical chemistry is nitric acid eating through wood.

3 0
1 year ago
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