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2 years ago

A strand of DNA has the following string of bases:

Chemistry

1 answer:

Ganezh [65]2 years ago

5 0

Answer:

ok so first of all try working hard then ill answer u

Explanation:

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QUESTION: What is the main function (job) of the Nervous System?

Tresset [83]

Answer:

It is C

Explanation:

it sends signals to the brain so it would be C

8 0

2 years ago

Read 2 more answers

Does the volume of a liquid change

nydimaria [60]

Answer:

Yes it does change

Explanation:

6 0

3 years ago

Identify the control group the experimental group the independent variable and the dependent variable

denis-greek [22]

Control group: 50 dogs continuing their normal diet
Experiments group: 50 dogs chosen to eat the new food
Independent variable: dog food
Dependent variable: the dogs’ weight

3 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

A mixture of0.161 moles of C is reacted with 0.117 moles of O2 in a sealed, 10.0 L-vessel at 500.0 K, producing a mixture of CO

labwork [276]

Answer:

number of moles of CO2 is 0.054

number of moles of CO is 0.107

number of moles of O2 remaining is 0.01 mole

mole fraction of CO is 0.63

Explanation:

Firstly, we write the equation of reaction;

3C(s) +2O2(g) → CO2(g) +2CO(g)

Now, we proceed.

From the written equation, we can deduce that

3 mol C = 2 mol O2 = 1 mol CO2 = 2 mol CO

No of mol of C reacted = 0.161 mol

limiting reactant according to the question is Carbon

a. no of mol of CO2 formed = 0.161*1/3 = 0.054 moles ( no of moles of CO2 formed is one-third of no of moles of carbon reacted. This is obtainable from their mole ratio 1:3)

b. no of mol of CO formed = 0.161*2/3 = 0.107 mol

c. no of mol of O2 remaining = 0.117 - (0.151*2/3) = 0.117-0.107 = 0.01 mole

d. mole fraction of CO = no of mol of CO/Total number of moles

= 0.107/(0.107+0.054+0.01)

= 0.625730994152 which is approximately 0.63

5 0

3 years ago