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3 years ago

Which statement about the atomic nucleus is correct

Chemistry

1 answer:

Sladkaya [172]3 years ago

8 0

The nucleus is made of protons and neutrons. It has a positive charge.

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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;

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

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r

rodikova [14]

Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

= 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

$K=Ae^{\frac{-E}{RT}}$

If $K_1=Ae^{\frac{-E_1}{RT}} -------equation 1$ &

$K_2=Ae^{\frac{-E_2}{RT}} -------equation 2$

$\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}$

$E_1= E_2-RT*In(\frac{K_1}{K_2})$

$E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})$

$E_1 = 28957.39292$ $J/mol$

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

8 0

3 years ago

How many liters of dinitrogen tetoxide are formed from 2.5 L of nitrogen?

Black_prince [1.1K]

Answer:

2.5 L

Explanation:

The stoichiometric ratio of N2 to N2O4 is 1:1

7 0

3 years ago

What changes if the degree (amount) of intermolecular bonds of a substance are changed?

elixir [45]

<h2>Answer:</h2>

The correct answers are;

volume of a substance

density of a substance

The volume and the density of the substance depend on inter molecular forces of the substance. So, when we will change the degree of inter molecular forces, whether decrease it or increase it, it will affect the volume and density of a substance.

7 0

3 years ago

Read 2 more answers

What is the standard notation for 7.934 x 10-4 ?

maks197457 [2]

Answer:

It is

Explanation:

75.34 Im hopeing this correct.Very sorry if wrong.

4 0

3 years ago