All categories

3 years ago

Before starting a spreadsheet, it is smarter to plan ahead and think through the design. true or false

1 answer:

6 0

True. THe axis order (horizontal/vertical) for instance, can cause a serious headache to you when organized wrong.
The same goes with functions, it's better to check them mentally/in a piece of paper beforehand.

You might be interested in

In the last two decades, how have cameras evolved?

Artist 52 [7]

Answer:

The Changes to Camera Technology over the last 70 years. The technology that the modern camera is based on was created several hundred years ago. ... The company would then reload the camera with new film and send it back to the consumer so they could take more pictures with it.

Explanation:

8 0

3 years ago

Read 2 more answers

Create a program in c/c++ which accepts user input of a decimal number in the range of 1 -100. Each binary bit of this number wi

kirill115 [55]

Answer:

// here is code in C++.

#include <bits/stdc++.h>

using namespace std;

void switch_fun(int input)

{

//array to store binary of input number

int bin[7];

// convert the number into binary

for(int i=0; input>0; i++)

{

bin[i]=input%2;

input= input/2;

}

// print the switch number and their status

cout<<"Switch number:\t 1\t 2\t 3\t 4\t 5\t 6\t 7 \n";

cout<<"status: ";

for(int j=6;j>=0;j--)

{

// if bit is 1 print "ON" else print "OFF"

if(bin[j]==1)

cout<<"\t ON";

else

cout<<"\t OFF";

}

int main()

{

int n;

cout<<"enter the number between 1-100 only:");

// read a number in decimal

cin>>n

// validate the input

while(n<1 ||n>100)

{

cout<<"wrong input!! Input must be in between 1-100:"<<endl;

cout<<"enter the number again :";

cin>>n;

}

// call the function with input parameter

switch_fun(n);

return 0;

}

Explanation:

Read a decimal number from user. If the number is not between 1 to 100 the it will again ask user to enter the number until the input is in between 1-100.Then it will call the fun() with parameter n. First it will convert the decimal to binary array of size 7.Then print the status of switch.If bit is 1 then print "ON" else it will print "OFF".

Output:

enter the number between 1-100 only:-5

wrong input!! Input must be in between 1-100:

enter the number again :125

wrong input!! Input must be in between 1-100:

enter the number again :45

Switch number: 1 2 3 4 5 6 7

Status: OFF ON OFF ON ON OFF ON

8 0

3 years ago

An argument does not always have to be made in words. A piece of music

mixas84 [53]

Answer:

B. emotional

Explanation:

An emotional argument. An argument does not always have to be made in words.

4 0

3 years ago

In media literacy,

Bas_tet [7]

what is mean by computers

7 0

2 years ago

What is the rate constant at 25.0 ∘c based on the data collected for trial b?

Alecsey [184]

<h2><u>Answer: </u></h2>

acetone + I2 + HCl ---> iodated acetone

Equation:

rate = k * [acetone]^x * [I2]^y * [HCl]^z

Once we know x, y, z, we can plug in any of the trials A->D and determine k NOTE: We can't use run E because temperature has an effect on rate. E was run at a different temperature.

The first thing to note is that do NOT have concentrations. We have volumes at a given molarity.

Table:

.001M I2.. ..050M HCl.. .1.0M acetone.. .water.. temp..time.. total vol

.. ....mL.. ... ... .. .mL.. .. ... .. .. mL.. .. .. .. ..mL.. ..°C.. .sec.. .. .. .L

A.. ...5.. .. .. .. .. ..10.. .. .. .. .. ..10.. .. .. .. ..25.. .. .25.. .130.. .. 0.05

B.. ..10.. .. .. ... .. 10.. .. .. ... .. .10.. .. .. ... ..20.. . .25.... 249.. ..0.05

C.. . 10.. .. .. .. .. .20.. .. .. .... .. 10.. .. ... ... .10... ..25.. ..128... .0.05

D.. . 10.. ... .. ... ..10.. .. ... .. ... 20.. .. .. .. .. 10.. .. 25.. ..131.. ..0.05

E.. ..10.. ... .. ... ..10.. ... ... ... ..10.. ... .. ... .20.. .. 42.8.. .38.. ..0.05

We can translate that into molarity in solution using this formula. (molarity pure ingredient * mL used / 1000 / total volume in liters)

.. .. ..I2.. .... HCl.. acetone.. temp.. ..rxn time

.. .. ..M.. .. ...M.. .... .M.. .... ..°C.. .. .. sec

A.. 0.0001.. 0.01..... 0.2.. .... .25... .... .130

B.. 0.0002.. 0.01.. .. 0.2.. .. .. 25.. .. .. .249

C.. 0.0002.. 0.02.. .. 0.2.. .. .. 25.. .. .. .128

D.. 0.0002.. 0.01.. .. 0.4.. .. . .25.. ... ...131

E.. 0.0002.. 0.01.. .. 0.2.. . ....42.8.. .. .. 38

From runs B and D, we can see that rate dropped by half .

As [I2] and [HCl] were held constant and [acetone] was doubled.

This means x=-1 in this equation

Rate = k * [acetone]⁻¹ * [I2]^y * [HCl]^z

Rate = k * [I2]^y * [HCl]^z

From runs A and B

[I2] doubles

[HCl] remains the same

[acetone] remains the same

rate doubles as [I2] doubles, rate doubles

y = 1 rate = k * [acetone]⁻¹ * [I2]¹ * [HCl]^z

And from runs B and C, we can see that , As [HC] doubles, (all else equal) the rate halves.

Z = -1

Rate = k * [I2] / ([acetone] * [HCl])

Rearranging

k = rate * [acetone] * [HCl] / [I2]

From any experimental run (A-D), we can calculate k.

using A to calc k... ..k = 2600 M²/sec

using B to calc k... . k = 2490 M²/sec

using C to calc k... . k = 2560 M²/sec

using D to calc k... . k = 2620 M²/sec

NOTE.. the problem statement said to use the data from run B to calc k.

Hence Final Answer:

k = 2490 M²/sec

4 0

3 years ago