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ivolga24 [154]
3 years ago
5

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 10

5 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 31 s for 1.0 L of O2 gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of 1.0 L; in other words, rate is the amount that diffuses over the time it takes to diffuse.)

Chemistry
1 answer:
aivan3 [116]3 years ago
6 0

Answer: the molar Mass of the unknown gas is 367.12

Explanation:Please see attachment for explanation

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A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som
Andrews [41]

The given question is incomplete. The complete question is as follows.

A certain liquid has a normal boiling point of 124.2^{o}C and a boiling point elevation constant k_{b} = 0.62 ^{o}C kg mol^{-1}. A solution is prepared by dissolving some sodium chloride (NaCl) in 6.50 g of X. This solution boils at 127.4^{o}C. Calculate the mass of NaCl that was dissolved. Round your answer to significant digits.

Explanation:

As per the colligative property, the elevation in boiling point will be as follows.

     

T = boiling point of the solution =

T_{o} = boiling point of the pure solvent = 124.2^{o}C

K_{b} = elevation of boiling constant = 0.62 ^{o}C kg mol^{-1}

We will calculate the molality as follows.

     molality = \frac{\text{maas of solute}}{\text{molar mass of solute}} \times \frac{1000}{\text{mass of solvent in g}}

i = vant hoff's factor

As NaCl is soluble in water and dissociates into sodium and chlorine ions so i = 2.

Putting the given values into the above formula as follows.  

    T - T_{o} = i \times K_{b} \times \text{molality of solution}

  (127.4 - 124.2)^{o}C = 2 \times 0.62 \times \frac{m}{60} \times \frac{1000}{650}

                  m = 100 g

Therefore, we can conclude that 100 g of NaCl was dissolved.

3 0
3 years ago
g Suppose 0.0350 g M g is reacted with 10.00 mL of 6 M H C l to produce aqueous magnesium chloride and hydrogen gas. M g ( s ) +
iren2701 [21]

Answer:

Mg will be the limiting reagent.

Explanation:

The balanced reaction is:

Mg + 2 HCl → MgCl₂ + H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Mg: 1 mole
  • HCl: 2 moles
  • MgCl₂: 1 mole
  • H₂: 1 mole

Being the molar mass of each compound:

  • Mg: 24.3 g/mole
  • HCl: 36.45 g/mole
  • MgCl₂: 95.2 g/mole
  • H₂: 2 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Mg: 1 mole* 24.3 g/mole= 24.3 g
  • HCl: 2 moles* 36.45 g/mole= 72.9 g
  • MgCl₂: 1 mole* 95.2 g/mole= 95.2 g
  • H₂: 1 mole* 2 g/mole= 2 g

0.0350 g of Mg is reacted with 10.00 mL (equal to 0.01 L) of 6 M HCl.

Molarity being the number of moles of solute that are dissolved in a certain volume, expressed as:

Molarity=\frac{number of moles of solute}{volume}

in units \frac{moles}{liter}

then, the number of moles of HCl that react is:

6 M=\frac{number of moles of HCl}{0.01 L}

number of moles of HCl= 6 M*0.01 L

number of moles of HCl= 0.06 moles

Then you can apply the following rule of three: if by stoichiometry 2 moles of HCl react with 24.3 grams of Mg, 0.06 moles of HCl react with how much mass of Mg?

mass of Mg=\frac{0.06 moles of HCl* 24.3 grams of Mg}{2 moles of HCl}

mass of Mg= 0.729 grams

But 0.729 grams of Mg are not available, 0.0350 grams are available. Since you have less mass than you need to react with 0.06 moles of HCl, <u><em>Mg will be the limiting reagent.</em></u>

7 0
3 years ago
Look at the graph
Vlad [161]

Answer:

a item dropping

Explanation:

KE is movement and PE is height. As it's falling it gets KE and falls downward giving it more KE and less PE

7 0
3 years ago
When sodium atoms (Na) and chlorine atoms (CT) join to make sodium chloride, or table salt, they form an ionic bond, Using this
jasenka [17]

Answer:

D I'm pretty sure.

Explanation:

I'm taking the quiz right now and D makes the most sense.

3 0
3 years ago
Read 2 more answers
Consider the reaction. 4 k(s) + o2(g) s 2 k2o(s) the molar mass of k is 39.09 g&gt;mol and that of o 32.00 g&gt;mol. Without doi
Scilla [17]

Answer : The correct option is (d) 1.5 g K, 0.38 g O₂

Explanation :

The molar masses of potassium and oxygen are very close to each other. Therefore we can assume them to be equal. If we assume that, then according to reaction stoichiometry, 4 moles of K are needed to react with 1 mol of O₂. Since the molar masses are assumed to be equal , we can say that the mass of potassium needed to react with that of oxygen should be 4 times the mass of oxygen.

From the given options, the only option that has less amount of K is option d.

Here, 0.38 g of oxygen needs 0.38 x 4 = 1.52 g of K. But the given mass of potassium is 1.5 g which is less. This indicates that potassium is the limiting reactant as we do not have enough potassium to completely react with all of the oxygen.

Therefore option d is the correct option

3 0
3 years ago
Read 2 more answers
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