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3 years ago

Based on the trend in the table, hypothesize what the volume of the sample at 400K would be.

Chemistry

1 answer:

mariarad [96]3 years ago

6 0

Answer: The volume of sample at 400 K is $285cm^3$

Explanation:

Charles' Law: This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

$V\propto T$ (At constant pressure and number of moles)

${\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1$ = initial volume of gas = $257cm^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = 360 K

$T_2$ = final temperature of gas = 400 K

Putting in the values we get:

${\frac{257}{360}=\frac{V_2}{400}$

$V_2=285cm^3$

Thus volume of sample at 400 K is $285cm^3$

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Describe the experimental basis for believing that the nucleus occupies a very small fraction of the volume of the atom.

svetlana [45]

Explanation:

Rutherford conducted an experiment in which he took a thin gold particle film on which he passes alpha- particles. He noticed that:

Most of the alpha particles get through the film and can be detected by the detector.
Around small portion of the alpha particle deflected at small angles.
A very very few alpha particle (approximately 1 out of 1 million alpha particles) just retraced their path which means come back from the center.

He concluded that:

Most of the space of the atom is empty and in the center of the atom , there is solid mass which is the cause of the alpha particles to come back. He gave the term nucleus to this solid mass.

4 0

4 years ago

Which of the following questions is outside the scope of science?

Airida [17]

A. How religion and philosophy different
All the other choices (b-d) can be explained by science while a can be explain by history I think

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3 years ago

Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compou

Ulleksa [173]

Answer:

$4.36~g~XY$

Explanation:

In this case, we can start with the reaction:

$2X + Y_2~->~2XY$

If we check the reaction, we will have 2 X and Y atoms on both sides. So, the reaction is balanced. Now, the problem give to us two amounts of reagents. Therefore, we have to find the limiting reagent. The first step then is to find the moles of each compound using the molar mass:

$3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X$

$4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2$

Now, we can divide by the coefficient of each compound (given by the balanced reaction):

$\frac{0.04~mol~X}{1}=~0.04$

$\frac{0.0875~mol~Y_2}{2}=0.04375$

The smallest value is for "X", therefore this is our limiting reagent. Now, if we use the molar ratio between "X" and "XY" we can calculate the moles of XY, so:

$0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY$

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol $Y_2$ = 48 g $Y_2$ (therefore 1 mol Y = 24 g Y). With this in mind the molar mass of XY would be 85+24 = 109 g/mol. With this in mind:

$0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY$

I hope it helps!

6 0

3 years ago

Determine the theoretical yield of HCl if 60.0 g of BC13 and 37.5 g of H20 are reacted according to the following balanced react

Pie

Answer : The theoretical yield of HCl is, 56.1735 grams

Explanation : Given,

Mass of $BCl_3$ = 60 g

Mass of $H_2O$ = 37.5 g

Molar mass of $BCl_3$ = 117 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $HCl$ = 36.5 g/mole

First we have to calculate the moles of $BCl_3$ and $H_2O$ .

$\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{60g}{117g/mole}=0.513moles$

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{37.5g}{18g/mole}=2.083moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$BCl_3(g)+3H_2O(l)\rightarrow H_3BO_3(s)+3HCl(g)$

From the balanced reaction we conclude that

As, 1 mole of $BCl_3$ react with 3 mole of $H_2O$

So, 0.513 moles of $BCl_3$ react with $3\times 0.513=1.539$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $BCl_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $HCl$ .

As, 1 mole of $BCl_3$ react to give 3 moles of $HCl$

So, 0.513 moles of $BCl_3$ react to give $3\times 0.513=1.539$ moles of $HCl$

Now we have to calculate the mass of $HCl$ .

$\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl$