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Marta_Voda [28]
3 years ago
9

How many moles in 68.1 grams of Cu(OH)2

Chemistry
1 answer:
Andrew [12]3 years ago
6 0

To do this, you would first add together the molar mass of all involved elements, to find how many grams are in a mole of Cu(OH)2. Keep in mind, the molar mass is equal to the atomic mass of an element in grams. For example the molar mass of copper (Cu) would be 63.55 (with 2 sig. figs.)

Therefore, now we add together the mass of all elements involved.

Cu: (63.55)+O2(15.99x2=31.98)+H2(1.01x2=2.02)

63.55+31.98+2.02= 97.55g per mole of Cu(OH)2.

Now, divide what we have by how much it takes to get a mole of the stuff.

68.1/97.55= 0.698mol Cu(OH)2


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Given that a for HBrO is 2. 8×10^−9 at 25°C. What is the value of b for BrO− at 25°C?
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If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).

<h3>What is base dissociation constant? </h3>

The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.

The dissociation reaction of hydrogen cyanide can be given as

HCN --- (H+) + (CN-)

Given,

The value of Ka for HCN is 2.8× 10^(-9)

The correlation between base dissociation constant and acid dissociation constant is

Kw = Ka × Kb

Kw = 10^(-14)

Substituting values of Ka and Kw,

Kb = 10^(-14) /{2.8×10^(-9) }

= 3.5× 10^(-6)

Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).

DISCLAIMER: The above question have mistake. The correct question is given as

Question:

Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?

learn more about base dissociation constant:

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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;

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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.

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