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Sliva [168]
2 years ago
15

A 0.5 mol sample of He(g) and a 0.5 mol sample of Ne(g) are placed separately in two 10.0 L rigid containers at 25°C. Each conta

iner has a pinhole opening. Which of the gases, He(g) or Ne(g), will escape faster through the pinhole and why?
A) He because the He atoms are moving at a higher average speed than the Ne atoms.
B) Ne because its initial pressure in the container is higher
C) Ne because the Ne atoms have a higher average kinetic energy than the He atoms
D) Both gases will escape at the same rate because the atoms of both gases have the same average kinetic enery
Chemistry
1 answer:
valentinak56 [21]2 years ago
5 0

Answer:

A) He because the He atoms are moving at a higher average speed than the Ne atoms.

Explanation:

The expression for the mean speed is:

C_{avg}=\sqrt {\dfrac{8RT}{\pi M}}

R is Gas constant having value = 8.314 J / K mol  

M is the molar mass of gas

T is the temperature

Thus, as seen from the formula,  

The average speed of the gas is inversely proportional to the square root of the molar mass of the gas

So, Helium has lower molecular mass than neon and thus, The speed of the sound in helium is greater when compared to that of neon.

<u>Thus, He will escape faster through the pinhole.</u>

<u>Option A. is correct.</u>

You might be interested in
What is the maximum number of hydrogen atoms that can be covalently bonded in a molecule containing two carbon atoms?a) 2b) 3c)
aksik [14]

Answer:

           The correct answer is Option-D (6 Hydrogen atoms).

Explanation:

                  Carbon atom has a unique property of linking to it self and making a chain of carbon called as Catenation. In a molecule having two carbon atoms there must be a bond between the two carbon atoms. The bond can be saturated (single bond) or unsaturated (double or triple bond). But as the statement states maximum number of hydrogen atoms so, we will asume the bond between two carbon atoms to be single.

                  As carbon has four valence electrons so, it has the ability to make four single bonds. In given molecule each carbon has already made a single bond with another carbon therefore, each carbon is left with three more unpaired electrons which will make covalent bond with three hydrogen atoms each as shown in attached structure.

8 0
2 years ago
50 kg of N2 gas and 10kg of H2 gas are mixed to produce NH3 gas calculate the NH3gas formed. Identify the limiting reagent in th
statuscvo [17]

Answer:

1. H2 is the limiting reactant.

2. 56666.67g ( i.e 56.67kg) of NH3 is produced.

Explanation:

Step 1:

The equation for the reaction. This is given below:

N2 + H2 —> NH3

Step 2:

Balancing the equation.

N2 + H2 —> NH3

The above equation can be balanced as follow :

There are 2 atoms of N on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NH3 as shown below:

N2 + H2 —> 2NH3

There are 6 atoms of H on the right side and 2 atoms on the left side. It can be balance by putting 3 in front of H2 as shown below

N2 + 3H2 —> 2NH3

Now the equation is balanced.

Step 3:

Determination of the masses of N2 and H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:

N2 + 3H2 —> 2NH3

Molar Mass of N2 = 2x14 = 28g/mol

Molar Mass of H2 = 2x1 = 2g/mol

Mass of H2 from the balanced equation = 3 x 2 = 6g

Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

Mass of NH3 from the balanced equation = 2 x 17 = 34g

From the balanced equation above,

28g of N2 reacted with 6g of H2 to produce 34g of NH3

Step 4:

Determination of the limiting reactant. This is illustrated below:

N2 + 3H2 —> 2NH3

Let us consider using all the 10kg (i.e 10000g) of H2 to see if there will be any left of for N2.

From the balanced equation above,

28g of N2 reacted with 6g of H2.

Therefore, Xg of N2 will react with 10000g of H2 i.e

Xg of N2 = (28 x 10000)/6

Xg of N2 = 46666.67g

We can see from the calculations above that there are leftover for N2 as only 46666.67g reacted out of 50kg ( i.e 50000g) that was given. Therefore, H2 is the limiting reactant.

Step 5:

Determination of the mass of NH3 produced during the reaction. This is illustrated below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

6g of H2 reacted to produce 34g of NH3.

Therefore, 10000g of H2 will react to produce = ( 10000 x 34)/6 = 6g of 56666.67g of NH3.

Therefore, 56666.67g ( i.e 56.67kg) of NH3 is produced.

3 0
2 years ago
Read 2 more answers
"The compound K2O2 also exists. A chemist can determine the mass of K in a sample of known mass that consists of either pure K2O
o-na [289]

Answer:

Yes, the chemist can determine which compound is in the sample.

Explanation:

In 1 mole of K₂O, the mass of K is 2 × 39.1 g = 78.2 g and the mass of K₂O is 94.2 g. The mass ratio of K to K₂O is 78.2 g / 94.2 g = 0.830.

In 1 mole of K₂O₂, the mass of K is 2 × 39.1 g = 78.2 g and the mass of K₂O₂ is 110.2 g. The mass ratio of K to K₂O₂ is 78.2 g / 110.2 g = 0.710.

If the chemist knows the mass of K and the mass of the sample, he or she must calculate the mass ratio of K to the sample.

  • If the ratio is 0.830, the compound is pure K₂O.
  • If the ratio is 0.710, the compound is pure K₂O₂.
  • If the ratio is not 0.830 or 0.710, the sample is a mixture.
6 0
2 years ago
What is the reaction energy Q of this reaction? Use c2=931.5MeV/u. Express your answer in millions of electron volts to three si
emmasim [6.3K]

Answer:

Energy= 2.7758 × 10^-11 J ;

71.112×10^-6 kJ.

Mass defect in Kilogram= 3.0885×10^-28 kg.

That is; 3.1×10^-28 kg(to two significant figure).

Explanation:

(Note: Check equation of reaction in the attached file/picture).

STEP ONE: we have to calculate the Mass defect.

Mass defect= Mass of reactants -- Mass of products.

Mass of the products: (140.9144+91.9262+3.060) u.

= 235.8666 u.

Mass of reactants: (1.0087+235.0439) u= 236.0526 u.

Therefore, the Mass defect= (236.0526 -- 235.8666) u

= 0.1860 u.

STEP TWO: Converting the Mass defect to energy;

0.18860 × 1.6605 × 10^27 kg

= 3.0885× 10^-28 kg

STEP THREE: Calculating energy released . Recall(from the question) c^2= 931.5 Mev/u. This is also equals to 9×10^16 m/s.

E=Mc^2.

Where E= energy released, c= speed of light, M= Mass.

Slotting in the values;

E= 3.0885×10^-28 kg × 9×10^16 m/s.

E=2.7758 × 10^-11 J.

Know that;( 1g of uranium × 1 mol of uranium ÷ 235.0439 g of uranium) × (6.002×10^23 atom of uranium/ 1 mol of uranium) × 2.7758× 10^-11.

=7.1112×10^-10 J

= 71.112×10^6 kJ.

3 0
2 years ago
Hydrogen was collected over water using the approach in the manual. The water temperature was 220C and the measured pressure ins
olasank [31]

Answer:

Pressure of hydrogen gas = 695.2 mmHg

Explanation:

Given:

Water temperature = 22°C

Pressure inside the tube = 715 mmHg

Find:

Pressure of hydrogen gas

Computation:

Using vapor pressure of water table

Water pressure at 22°C = 19.8 mmHg

Pressure inside the tube = Pressure of hydrogen gas + Water pressure at 22°C

715 = Pressure of hydrogen gas + 19.8

Pressure of hydrogen gas = 715 - 19.8

Pressure of hydrogen gas = 695.2 mmHg

3 0
2 years ago
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