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Elza [17]
3 years ago
7

If 3.0 liters of oxygen gas react with excess carbon monoxide at STP, how many liters of carbon dioxide can be produced under th

e same conditions?
2 CO (g) + O2 (g) ---> 2 CO2 (g)

a. 1.5 L
b. 3.0 L
c. 4.5 L
d. 6.0 L,
Chemistry
1 answer:
AlladinOne [14]3 years ago
6 0
At STP, the volume of a gas represents the number of particles.That said, from the chemical reaction one mole of oxygen reacts with two moles of co to produce the product, CO2At STP, 3 moles of Oxygen will produce 6 moles of CO2. Hence It follows that at standard temperature and pressure, 6.0 L of CO2 will be produced. Option D.
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How to find the relationship between slope and speed
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You can study article about slope and speed 
3 0
3 years ago
A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 degrees celsius,
Airida [17]

Answer:

0.714 liter.

Explanation:

Given:

The balloon initially has a volume of 0.4 liters and a temperature of 20 degrees Celsius.

It is heated to a temperature of 250 degrees Celsius.

Question asked:

What will be the volume of the balloon after he heats it to a temperature of 250 degrees Celsius ?

Solution:

By using:

PV=nRT

Assuming pressure as constant,

V∝ T

Now, let  K is the constant.

V = KT

Let initial volume of balloon , V_{1} = 0.4 liter

1000 liter = 1 meter cube

1 liter = \frac{1}{1000} m^{3} = 10^{-3} m^{3

0.4 liter = 0.4\times10^{-3}=4\times10^{-4} m^{3}

And initial temperature of balloon, T_{1} = 20°C = (273 + 20)K

                                                                          = 293 K

Let the final volume of balloon is V_{2}

And a given, final temperature of balloon, T_{2} is 250°C = (273 + 250)K

                                                                                          = 523 K

Now, V_{1} = KT_{1}

          4\times10^{-4}=K\times293\ (equation\ 1 )

V_{2} = KT_{2}

    =K\times523\ (equation 2)

Dividing equation 1 and 2,

 \frac{4\times10^{-4}}{V_{2} } =\frac{K\times293}{K\times523}

K cancelled by K.

By cross multiplication:

293V_{2} =4\times10^{-4} \times523\\V_{2} =\frac{ 4\times10^{-4} \times523\\}{293} \\          = \frac{2092\times10^{-4}}{293} \\          =7.14\times10^{-4}m^{3}

Now convert it into liter with the help of calculation done above.

7.14\times10^{-4} \times1000\\7.14\times10^{-4} \times10^{3} \\0.714\ liter

Therefore, the volume of the balloon be after he heats it to a temperature of 250 degrees Celsius is 0.714 liter.

5 0
3 years ago
Use the successive ionization energies for this unknown element to identify the family it belongs to.
NNADVOKAT [17]

Answer: Belongs to the group 2A

Explanation:

As you can see, the first two ionization energies are close and low, meaning that this element ionizates easily.

Not only loses easily the first electron, but the second too

To remove the third electron you requiered a huge amount of energy

Now, elements easily ionizable are the ones from group IA, group 2A and transition metals.

The last ones have mixed characteristics in matter of how many electrons you can remove from them, so they are not a family.

Now the question: group I or group II ?

The elements of group I have low  ionization energies  for the first electron but high energies for the second ones.

Being all that said, the unknown element belongs to the Group 2A

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3 years ago
In which color coded area would you find an element that behaves both like a metal and a non-metal?
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what elements are each color over? can you list one from each color so i can answer it

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3 years ago
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Last one cuz it stay in the air for a while then swings making it kinetic
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