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statuscvo [17]
2 years ago
6

Hexadecimal to denary gcse method

Computers and Technology
1 answer:
Anuta_ua [19.1K]2 years ago
3 0

There are two ways to convert from hexadecimal to denary gcse method. They are:

  • Conversion from hex to denary via binary.
  • The use of base 16 place-value columns.

<h3>How is the conversion done?</h3>

In Conversion from hex to denary via binary:

One has to Separate the hex digits to be able to know or find its equivalent in binary, and then the person will then put them back together.

Example - Find out the denary value of hex value 2D.

It will be:

2 = 0010

D = 1101

Put them them together and then you will have:

00101101

Which is known to be:

0 *128 + 0 * 64 + 1 *32 + 0 * 16 + 1 *8 + 1 *4 + 0 *2 + 1 *1

= 45 in denary form.

Learn more about hexadecimal from

brainly.com/question/11109762

#SPJ1

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Read 2 more answers
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
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