Linda is learning how to bowl and

(I NEED THIS ANSWERED QUICKLY!)

kiruha [24]

Answer:

C. distance between two of the numbered lines

4 0

3 years ago

Read 2 more answers

Calculate the amount of heat released in the combustion of 10.5 grams of Al with 3 grams of O2 to form Al2O3(s) at 25°C and 1 at

ElenaW [278]

Answer : The amount of heat released in the combustion is, 209.5 kJ

Explanation :

First we have to calculate the moles of Al and $O_2$ .

$\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{10.5g}{27g/mole}=0.389moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{3g}{32g/mole}=0.188moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$4Al+3O_2\rightarrow 2Al_2O_3$

From the balanced reaction we conclude that

As, 3 mole of $O_2$ react with 4 mole of $Al$

So, 0.188 moles of $O_2$ react with $\frac{4}{3}\times 0.188=0.251$ moles of $Al$

From this we conclude that, $Al$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Al_2O_3$

From the reaction, we conclude that

As, 3 mole of $O_2$ react to give 2 mole of $Al_2O_3$

So, 0.188 moles of $O_2$ react to give $\frac{2}{3}\times 0.188=0.125$ moles of $Al_2O_3$

Now we have to calculate the amount of heat released in the combustion.

As, 1 mole of $Al_2O_3$ releases amount of heat = 1676 kJ

So, 0.125 mole of $Al_2O_3$ releases amount of heat = $0.125\times 1676kJ=209.5kJ$

Thus, the amount of heat released in the combustion is, 209.5 kJ

4 0

3 years ago

1.674×10-4 mol of an unidentified gaseous substance effuses through a tiny hole in 86.6 s. Under identical conditions, 1.715×10-

siniylev [52]

<h2>Answer:</h2>

44.06 g/mol

<h3>Explanation:</h3>

We are given;

Number of moles of unidentified gas as 1.674×10^-4 mol
Time of effusion of unidentified gas 86.6 s
Number of moles of Argon gas as 1.715×10^-4 mol
Time of effusion of Argon gas is 84.5 s

We are supposed to calculate the molar mass of unidentified gas

<h3>Step 1: Calculate the effusion rates of each gas</h3>

Effusion rate = Number of moles/time

Effusion rate of unidentified gas (R₁)

= 1.674×10^-4 mol ÷ 86.6 s

= 1.933 × 10^-6 mol/s

Effusion rate of Argon gas (R₂)

= 1.715×10^-4 mol ÷ 84.5 sec

= 2.030 × 10^-6 mol/s

<h3>Step 2: Calculate the molar mass of unidentified gas</h3>

Assuming the molar mass of unidentified gas is x;
We can use the Graham's law of effusion to find x;
According to Graham's law of diffusion;

$\frac{R_{1}}{R_{2}}}=\frac{\sqrt{MM_{Ar}}}{\sqrt{X}}$

But, Molar mass of Argon is 39.948 g/mol

Therefore;

$\frac{1.933*10^-6mol/s}{2.030*10^-6mol/s}}=\frac{\sqrt{39.948}}{\sqrt{X}}$

$0.9522=\frac{\sqrt{39.948}}{\sqrt{X}}$

Solving for X

x = 44.06 g/mol

Therefore, the molar mass of the identified gas is 44.06 g/mol

3 0

4 years ago

A sample of xenon gas occupies a volume of 6.80 L at 52.0°C and 1.05 atm. If it is desired to increase the volume of the gas sam

Pavlova-9 [17]

Answer:

207.03°C

Explanation:

The following data were obtained from the question:

V1 (initial volume) = 6.80 L

T1 (initial temperature) = 52.0°C = 52 + 273 = 325K

P1 (initial pressure) = 1.05 atm

V2 (final volume) = 7.87 L

P2 (final pressure) = 1.34 atm

T2(final temperature) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas sample can be obtained as follow:

P1V1/T1 = P2V2/T2

1.05 x 6.8/325 = 1.34 x 7.87/T2

Cross multiply to express in linear form as shown below:

1.05 x 6.8 x T2 = 325 x 1.34 x 7.87

Divide both side by 1.05 x 6.8

T2 = (325 x 1.34 x 7.87) /(1.05 x 6.8)

T2 = 480.03K

Now, let us convert 480.03K to a number in celsius scale. This is illustrated below:

°C = K - 273

°C = 480.03 - 273

°C = 207.03°C

Therefore, the final temperature of the gas will be 207.03°C

5 0

3 years ago

Draw the Lewis structure with the lowest formal charges for ClF2+ . Include nonzero formal charges and lone pair electrons in th

masya89 [10]

Answer:

See explanation

Explanation:

The formal charge is obtained from;

Formal Charge = Valence electrons on atom - [number of bonds - lone pair electrons]

The correct structure of ClF2+ is the structure attached to this answer (image obtained from quora) in which the formal charge on fluorine is zero and the formal charge on chlorine is + 1. This is the correct structure because the chlorine is more electronegative than fluorine as expected.

4 0

3 years ago