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ArbitrLikvidat [17]
3 years ago
5

Once two gas particles undergo an elastic collision, the particles

Chemistry
1 answer:
babunello [35]3 years ago
8 0
D. have not lose any kinetic energy I'm pretty sure
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1. If I have 45 L of He in a balloon at 25 degrees celsius and increase the temperature of the
Greeley [361]

Use Charles' Law: V1/T1 = V2/T2. We assume the pressure and mass of the helium is constant. The units for temperature must be in Kelvin to use this equation (x °C = x + 273.15 K).

We want to solve for the new volume after the temperature is increased from 25 °C (298.15 K) to 55 °C (328.15 K). Since the volume and temperature of a gas at a constant pressure are directly proportional to each other, we should expect the new volume of the balloon to be greater than the initial 45 L.

Rearranging Charles' Law to solve for V2, we get V2 = V1T2/T1.  

(45 L)(328.15 K)/(298.15 K) = 49.5 ≈ 50 L (if we're considering sig figs).

7 0
3 years ago
When solutions of silver nitrate and sodium chloride are mixed, silver chloride precipitates out of the solution according to th
kipiarov [429]

The concentration of the sodium chloride would be 0.082 M

<h3>Stoichiometric calculations</h3>

From the equation of the reaction, the ratio of AgCl produced to NaCl required is 1:1.

Mole of 46.6 g AgCl produced = 46.6/143.32 = 0.325 moles

Equivalent mole of NaCl = 0.325 moles.

Molarity of 0.325 moles, 3.95 L NaCl = mole/volume = 0.325/3.95 = 0.082 M

More on stoichiometric calculations can be found here: brainly.com/question/27287858

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3 0
2 years ago
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Look at the diagram below , which shows gas particles in a container. If the piston was lowered, so that the size of the contain
raketka [301]

The pressure of the gas is expected to increase in accordance to Boyle's law.

<h3>What is Boyle's law?</h3>

Boyle's law states that, the volume of a given mass of gas is inversely proportional to its pressure at constant temperature.

By implication, when the piston is lowered and the volume of the gas is decreased, the pressure of the gas is expected to increase in accordance to Boyle's law.

Learn more about Boyle's law: brainly.com/question/1437490

3 0
3 years ago
Use the limiting reagent to determine how many grams of Cu(OH)2 should precipitate out in the reaction - CuSo4(aq) + 2NaOH(aq) -
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I think there is a lack of information in the given problem above such as the grams of copper sulfate and sodium hydroxide that was used in the experiment. Kindly resubmit the question with the complete details so that we can help you. Thank you.

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3 years ago
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The air bags in cars are inflated when a collision triggers the explosive, highly exothermic decomposition of sodium azide (NaN3
Oksanka [162]

Answer : The mass of NaN_3 required is, 166.4 grams.

Explanation :

First we have to calculate the moles of nitrogen gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of N_2 gas = 1.00 atm

V = Volume of N_2 gas = 113 L

n = number of moles N_2 = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 85^oC=273+85=358K

Putting values in above equation, we get:

1.00atm\times 113L=n\times (0.0821L.atm/mol.K)\times 358K

n=3.84mol

Now we have to calculate the moles of sodium azide.

The balanced chemical reaction is,

2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)

From the balanced reaction we conclude that

As, 3 mole of N_2 produced from 2 mole of NaN_3

So, 3.84 moles of N_2 produced from \frac{2}{3}\times 3.84=2.56 moles of NaN_3

Now we have to calculate the mass of NaN_3

\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3

Molar mass of NaN_3 = 65 g/mole

\text{ Mass of }NaN_3=(2.56moles)\times (65g/mole)=166.4g

Therefore, the mass of NaN_3 required is, 166.4 grams.

3 0
3 years ago
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