Can someone help me with this?

7.7. A painter leans his back against a painted wall while looking into a 1m long mirror at the opposite end of a rectangular ro

Karolina [17]

Answer:

Correct answer is b) 2m

Explanation:

8 0

3 years ago

A solution is made by mixing equal masses of methanol, CH 4 O , and ethanol, C 2 H 6 O . Determine the mole fraction of each com

Fynjy0 [20]

Answer:

Mole fraction of $CH_4O$ = 0.58

Mole fraction of $C_2H_6O$ = 0.42

Explanation:

Let the mass of $CH_4O$ and $C_2H_6O$ = x g

Molar mass of $CH_4O$ = 33.035 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles_{CH_4O}= \frac{x\ g}{33.035\ g/mol}$

$Moles_{CH_4O}=\frac{x}{33.035}\ mol$

Molar mass of $C_2H_6O$ = 46.07 g/mol

Thus,

$Moles= \frac{x\ g}{46.07\ g/mol}$

$Moles_{C_2H_6O}=\frac{x}{46.07}\ mol$

So, according to definition of mole fraction:

$Mole\ fraction\ of\ CH_4O=\frac {n_{CH_4O}}{n_{CH_4O}+n_{C_2H_6O}}$

$Mole\ fraction\ of\ CH_4O=\frac{\frac{x}{33.035}}{\frac{x}{33.035}+\frac{x}{46.07}}=0.58$

Mole fraction of $C_2H_6O$ = 1 - 0.58 = 0.42

6 0

4 years ago

At 400 K oxalic acid decomposes according to the reaction:H2C2O4(g)→CO2(g)+HCOOH(g)In three separate experiments, the intial pre

padilas [110]

Answer:

v = 2,66x10⁻⁵ P[H₂C₂O₄]

Explanation:

For the reaction:

H₂C₂O₄(g) → CO₂(g) + HCOOH(g)

At t = 0, the initial pressure is just of H₂C₂O₄(g). At t= 20000 s, pressures will be:

H₂C₂O₄(g) = P₀ - x

CO₂(g) = x

HCOOH(g) = x

P at t=20000 is:

P₀ - x + x + x = P₀+x. That means P at t=20000s - P₀ = x

For 1st point:

x = 92,8-65,8 = 27

Pressure of H₂C₂O₄(g) at t=20000s: 65,8-27 = 38,8

2nd point:

x = 130-92,1 = 37,9

H₂C₂O₄(g): 92,1 - 37,9 = 54,2

3rd point:

x = 157-111 = 46

H₂C₂O₄(g): 111-46 = 65

Now, as the rate law is :

v = k P[H₂C₂O₄]

Based on integrated rate law, k is:

(- ln P[H₂C₂O₄] + ln P[H₂C₂O₄]₀) / t = k

1st point:

k = 2,64x10⁻⁵

2nd point:

k = 2,65x10⁻⁵

3rd point:

k = 2,68x10⁻⁵

The averrage of this values is:

k = 2,66x10⁻⁵

That means law is:

v = 2,66x10⁻⁵ P[H₂C₂O₄]

I hope it helps!

4 0

3 years ago

What would the formula be for this model above?

Nadusha1986 [10]

Probably CH(subscript)4... :) It's Methane

4 0

4 years ago

2. 2. Consider the following reaction occurring in a closed chemical system. Assume that this reaction is at equilibrium and tha

Alika [10]

1. This is a combustion reaction.

Combustion reactions can happen with the presence of O₂ gas. O₂ reacts with another element or compound and oxidize it. Here ethanol reacts with O₂ and produces CO₂ and H₂O as products. Combustion is also called as burning. 

2. Reaction will shift to right. 

If more CH₃CH₂OH is added to the system, then the amount of CH₃CH₂OH will increase. Then the equilibrium in the system will be broken. To make the equilibrium again, the added CH₃CH₂OH should be removed. To do that system will consume more CH₃CH₂OH to make products which helps to decrease the amount of ethanol. Hence, the reaction will shift to right.

3. The reaction will shift to right.

If the water is extracted from the system, the amount of water will decrease. That means the amount of products decrease. Then the system will try to gain equilibrium by increasing the water. To increase water the forward reaction should be enhanced. Hence, the reaction will shift to right.

4. The reaction will shift to right.

This is an exothermic reaction since it produces heat. If the produced heat is removed, then the system will be cold. To maintain the temperature, system has to increase the amount of heat produced. Then, the forward reaction should be enhanced. Hence, the reaction will shift to right.

5. The Le Chatelier's principle.

Le Chatelier's principle says if a condition changes in a system which was in an equilibrium state, the system will try to gain equilibrium by correcting the changed condition back to normal. Most of industries which make chemicals use this principle

3 0

3 years ago

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