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miskamm [114]
4 years ago
7

Suppose you have the following declaration.char[] nameList = new char[100];Which of the following range is valid for the index o

f the array nameList.(i) 1 through 100(ii) 0 through 1001. Both are invalid2. Only (i)3. None of these4. Only (ii)
Computers and Technology
1 answer:
VMariaS [17]4 years ago
4 0

4. Only (ii)

<u>Explanation:</u>

The declaration of the array can be of two types:

1. int a[100];

2. int[] a = new int[100];

The general thing about an array is that whenever we want to undergo traversal in an array, we always have to start from the 0th position as the size of the array may be a whole number (let us say 10). So, to undergo traversal in an array, we start from 0 to n-1 (in this case 9) such that it covers the size of the array.

The size of an array can be finite or infinite. The general rule is it starts from 0 to n-1 where n is the size of the array. In the above example, the range of the index of the array will be 0 through 100 and not 1 through 100.

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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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