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Sedbober [7]
3 years ago
10

What quaintly represents .500 mole at STP?

Chemistry
2 answers:
s2008m [1.1K]3 years ago
3 0
.500 moles is roughly
.5 * 6.022 * 10^23 = 3.011*10^23 atoms. this is independent of STP
if it’s wrong i’m sorry
BARSIC [14]3 years ago
3 0
The answer is 500 is roughly
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Which of the following correctly describes a mixture? (4 points) Group of answer choices The particles are chemically bonded tog
Zanzabum

Answer:

The particles have no set ratio for how to combine, and they can be separated by physical means.

Explanation:

5 0
3 years 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
Please answer ASAP!!
Alenkasestr [34]

Answer:

C.0.28 V  

Explanation:

Using the standard cell potential we can find the standard cell potential for a voltaic cell as follows:

The most positive potential is the potential that will be more easily reduced. The other reaction will be the oxidized one. That means for the reactions:

Cu²⁺ + 2e⁻ → Cu E° = 0.52V

Ag⁺ + 1e⁻ → Ag E° = 0.80V

As the Cu will be oxidized:

Cu → Cu²⁺ + 2e⁻

The cell potential is:

E°Cell = E°cathode(reduced) - E°cathode(oxidized)

E°cell = 0.80V - (0.52V)

E°cell = 1.32V

Right answer is:

<h3>C.0.28 V </h3>

<h3 />

4 0
3 years ago
Determine the molar solubility of AgBr in a solution containing 0.150 M NaBr. Ksp (AgBr) = 7.7 × 10-13.
lesya692 [45]

Answer:

Molar solubility of AgBr = 51.33 × 10⁻¹³

Explanation:

Given:

Amount of NaBr = 0.150 M

Ksp (AgBr) = 7.7 × 10⁻¹³

Find:

Molar solubility of AgBr

Computation:

Molar solubility of AgBr = Ksp (AgBr) / Amount of NaBr

Molar solubility of AgBr = 7.7 × 10⁻¹³ / 0.150

Molar solubility of AgBr = 51.33 × 10⁻¹³

5 0
3 years ago
How is the stability of a nucleus affected by changing the number of nuclear particles?
Svet_ta [14]

Answer:

Greatly affected.

Explanation:

The stability of a nucleus is greatly affected by changing the number of nuclear particles because changing the number of nuclear particles i.e. neutron or proton will leads to change the element. When we change the number of protons in an atom, we will change the atom from one element to a different element. Sometimes, when you add a proton to an element, the element will become radioactive in nature so we can conclude that changing number of nuclear particles impact the stability of nucleus of an atom.

5 0
3 years ago
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