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Sedaia [141]
4 years ago
8

Only 5 questions! All you have to do is check my answers...1. A flask contains four gases: CH4, O2, C2H5, and N2. When the stopp

er is removed, which gas will diffuse the fastest?
CH4 <---
O2
C2H5
N2

2. Which gas effuses 2.39 times slower than nitrogen gas?

O2
Cl2
Br2 <---
I2

3. What is the molecular mass of a gas that effuses three times faster than radon?

16 g/mol
25 g/mol <---
50 g/mol
67 g/mol,
Chemistry
1 answer:
jarptica [38.1K]4 years ago
3 0

1. CH4
2. Br2
3. 25 g/mol

Graham's law indicates that the rate of diffusion is inversely proportional to the square root of its molecular weight. So the lighter the gas, the more rapid the diffusion. With that in mind, let's look at your questions:

1. A flask contains four gases: CH4, O2, C2H5, and N2. When the stopper is removed, which gas will diffuse the fastest?The answer here will be the lightest of the 4 gasses. So let's look at the molar mass of each gas:
CH4 = 16.04 g/mol
O2 = 32.00 g/mol
C2H5 = 29.06
N2 = 28.01
And CH4 is the lightest and therefore the fastest.

2. Which gas effuses 2.39 times slower than nitrogen gas? O2 Cl2 Br2 <--- I2
Let's start with the square root of 28.01 which is 5.292447449.
Now multiply by 2.39 and square the result to get the molar mass of the
desired gas which is (5.29*2.39)^2 = 12.64^2 = 159.77Now let's look at the molar masses.
O2 = 32
Cl2 = 70.91
Br2 = 159.81
I2 = 253.81
And the match goes to Br2

3. What is the molecular mass of a gas that effuses three times faster than radon? 16 g/mol 25 g/mol <--- 50 g/mol 67 g/mol,The molar mass of radon is 222 g/mol. So let's get the square root, divide by
3, then square the result. So(sqrt(222)/3)^2 = (14.89966443/3)^2 = 4.966554809^2 = 24.66666667 g/mol.And of the available choices, 25 g/mol is the best match.
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5) Solid silver chloride, AgCl, precipitates when solution of silver nitrate, AgNO3, and iron (111) chloride,
DanielleElmas [232]

The chemical reaction between the reactants:

3 AgNO₃ (aq) + FeCl₃ (aq) → 3 AgCl (s) +  Fe(NO₃)₃ (aq)

Explanation:

We have the following chemical reaction:

3 AgNO₃ (aq) + FeCl₃ (aq) → 3 AgCl (s) +  Fe(NO₃)₃ (aq)

Complete ionic equation:

3 Ag⁺ (aq) + 3 NO₃⁻ (aq) + Fe³⁺ (aq) + 3 Cl⁻ (aq) → 3 AgCl (s) +   Fe³⁺ (aq) + 3 NO₃⁻  (aq)

We remove the spectator ions and we get the net ionic equation:

Ag⁺ (aq) + Cl⁻ (aq) → AgCl (s)

where:

(aq) - aqueous

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Learn more about:

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4 0
3 years ago
What is an example of equilibrium?
frez [133]

A good example of equilibrium would be the mixing of oil and water in a closed container.

<h3>What is chemical equilibrium?</h3>

Chemical equilibrium is a condition in which the concentrations of components of a chemical reaction remain unchanged and have no tendency to change.

Of all the options, the only one where the concentrations of the component reactants cannot change is a mixture containing oil and water in a closed container.

Oil and water are immiscible and thus, their concentrations remain constant.

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5 0
2 years ago
A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther
sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )

1. Calculate the thermometer readings after  0.5 min and 1 min

(a) After 0.5 min

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}

(b) After 1 min

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

  t = t - 1, because the cooling starts 1 min late

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0
3 years ago
How many grams of ethylene glycol (c2h6o2 must be added to 1.25 kg of water to produce a solution that freezes at -5.88 ∘c?
e-lub [12.9K]
The freezing point depression is calculated through the equation,
                                    ΔT = (kf)  x m 
where ΔT is the difference in temperature, kf is the freezing point depression constant (1.86°C/m), and m is the molality. Substituting the known values,
                                   5.88 = (1.86)(m)
m is equal to 3.16m

Recall that molality is calculated through the equation,
                                  molality = number of mols / kg of solvent
                                       number of mols = (3.16)(1.25) = 3.95 moles
Then, we multiply the calculated amount in moles with the molar mass of ethylene glycol and the answer would be 244.9 g.

6 0
3 years ago
A 7.74 L balloon is filled with water at 3.88 atm. If the balloon is squeezed into a 0.23 L beaker and does NOT burst, what is t
Vladimir [108]

Answer:

131 atm

Explanation:

To find the new pressure, you need to use Boyle's Law:

P₁V₁ = P₂V₂

In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the new pressure (P₂) by plugging the given values into equation and simplifying.

P₁ = 3.88 atm                       P₂ = ? atm

V₁ = 7.74 L                           V₂ = 0.23 L

P₁V₁ = P₂V₂                                                      <----- Boyle's Law

(3.88 atm)(7.74 L) = P₂(0.23 L)                       <----- Insert values

30.0312 = P₂(0.23 L)                                      <----- Simplify left side

131 = P₂                                                          <----- Divide both sides by 0.23

6 0
2 years ago
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