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zepelin [54]
3 years ago
11

In the reaction: C6H6 + O2 = CO2 + H2O If 26.2 grams of C6H6 is reacted in a 0.250-liter container and this reaction is carried

out at 395 K, what is the pressure inside the container?
Chemistry
1 answer:
Lelechka [254]3 years ago
6 0
<h3>Answer:</h3>

43.33 atm

<h3>Explanation:</h3>

We are given;

Mass of C₆H₆ = 26.2 g

Volume of the container = 0.25 L

Temperature = 395 K

We are required to calculate the pressure inside the container;

First, we calculate the number of moles of C₆H₆

Molar mass of C₆H₆ =  78.1118 g/mol.

But; Moles = mass ÷ Molar mass

Moles of  C₆H₆ = 26.2 g ÷  78.1118 g/mol.

                         = 0.335 moles C₆H₆

Second, we calculate the pressure, using the ideal gas equation;

Using the ideal gas equation, PV = nRT , Where R is the ideal gas constant, 0.082057 L.atm/mol.K

Therefore;

P = nRT ÷ V

  = (0.335 mol × 0.082057 × 395 K) ÷ 0.25 L

 = 43.433 atm

Therefore, the pressure inside the container is 43.33 atm

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How many moles of salt are in 13.8 g of sodium chloride?
Hoochie [10]

Answer: 0.24 moles

Explanation:

Molecular Mass of NaCl (23 + 35.5) = 58.5g

58.5g of Sodium Chloride -------> 1 mole of NaCl

∴ 13.8g of Sodium Chloride  ------>  1 ÷58.5 x  13.8 = 0.2358974  ≈    0.24moles

                                                         -  

3 0
3 years ago
Read 2 more answers
Billy the friendly Robot uses 50 N of force to lift a box 2 meters in the air. How much work did Billy perform?
nexus9112 [7]

Answer:

Work done, W = 100 J

Explanation:

We have, Billy the friendly Robot uses 50 N of force to lift a box 2 meters in the air.

It is required to find the work done by Billy.

Work done by an object is given in terms of force and displacement. The formula used to find the work done is given by :

W=Fd\\\\W=50\ N\times 2\ m\\\\W=100\ J

So, the work performed by Billy is 100 J.

4 0
3 years ago
Design and implement a program (name it SumValue) that reads three integers (say X, Y, and Z) and prints out their values on sep
rjkz [21]

Answer:

class sum (

public static void sumofvalue (int m, int n, int p)

{

System.out.println(m);

System.out.println(n);

System.out.println(p);

int SumValue=m+n+p;

System.out.println("Average="+Sumvalue/3);

}

)

Public class XYZ

(

public static void main(String [] args)

{

sum ob=new sum();

int X=3;

int X=4;

int X=5;

ob.sumofvalue(X,Y,Z);

int X=7;

int X=8;

int X=10;

ob.sumofvalue(X,Y,Z);

}

)

Explanation:

The above program is made in Java, in which first we have printed value in a separate line. After that, the average value of those three values has been printed according to the question.

The processing of the program is given below in detail

* The first one class named 'sum' has been created which contains the function to print individual value and the average of those three values.

* In seconds main class named 'XYZ', the object of that the above class had been created which call the method of the above class to perform functions.

* In the main class values are assigned to variables X, Y, Z.

3 0
3 years ago
How many moles of H2SO4 are present in 0.500 L of a 0.150 M H2SO4 solution?
bonufazy [111]

Answer: 0.075

Explanation:

(concentration in molarity)(volume in liter) = answer

0.15 mol/L *0.500L = 0.075 mol

3 0
3 years ago
-1
Phantasy [73]
Carbonation isn’t a force that causes such
7 0
3 years ago
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