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lys-0071 [83]
3 years ago
9

A sample of crude sodium iodide was analysed by the following rection.

Chemistry
1 answer:
amm18123 years ago
6 0

A sample of crude sodium iodide was analyzed by the following balanced reaction. The oxidation number of S in SO₄²⁻ is +6.

8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻

Let's consider the following unbalanced redox reaction.

I⁻ + SO₄²⁻ → I₂ + H₂S

  • The oxidation number of I goes from -1 (I⁻) to 0 (I₂) so it is oxidized.
  • The oxidation number of S goes from +6 (SO₄²⁻) to -2 (H₂S) so it is reduced.

The corresponding half-reactions are:

I⁻ → I₂

SO₄²⁻ → H₂S

We will perform the mass balance adding OH⁻ and H₂O where appropriate.

2 I⁻ → I₂

6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻

Then, we will perform the charge balance adding electrons where appropriate.

2 I⁻ → I₂ + 2 e⁻

8 e⁻ + 6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻

Finally, we will multiply the first half-reaction by 4 and the second by 1, and add them.

4 × (2 I⁻ → I₂ + 2 e⁻)

1 × (8 e⁻ + 6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻)

------------------------------------------------------------

8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻

A sample of crude sodium iodide was analyzed by the following balanced reaction. The oxidation number of S in SO₄²⁻ is +6.

8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻

Learn more: brainly.com/question/2671074

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Compute the sugar content in an 8 oz sample of a soft drink. If the sugar content as per label on the product =10g per 100ml.​
Nataly_w [17]

Answer:

m_{sugar}=23.7g\ sugar

Explanation:

Hello,

In this case, we can first compute the volume of the sample in mL from the ounces:

8oz*\frac{29.5735mL}{1oz} =236.6mL

Thus, with the volume of the sample, we can compute the amount of sugar given the 10 g of sugar per 100 mL of soft drink as shown below:

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Determine the location of the last significant place value by placing a bar over the digit.
Vlad1618 [11]

Answer:

8,040  

0.0300  

699.5  

2.000 x 102

0.90100  

90, 100  

4.7 x 10-8  

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3.01 x 1021

0.000410

Explanation:

First remember the following rules of determining the last significant place value :

1. The digits from 1-9 are all significant and zeros between significant digits are also significant.

2.  The trailing or ending zeroes are significant only in case of a decimal number otherwise they are ignored. However starting zeroes of such a number are not significant.

Now observing above rules, lets determine the location of the last significant place value of each given example. I am determining the location by turning the last significant place to bold.

1) 8,040

8,040

Location of the last significant place value is 3 and bar is over last significant digit that is 4. Number is not decimal so ending zero is ignored. Every non zero digit is a significant.

2)  0.0300

0.0300

Location is 3 and bar is over 0. Number has a decimal point so ending zero is not ignored but starting zeroes are ignored.

3) 699.5

699.5

Location is 4 and bar is over 5.

4) 2.000 x 10²

2.000 x 10²

Location is 4 and bar is over 0. This is because the number is decimal so trailing zeroes cannot be ignored. Also if we convert this number it becomes:

200.0 so last significant digit is 0 and location of last significant digit is 4.

5) 0.90100

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Location is 5 and bar is over 0. This is because in a number with decimal point starting zeroes are ignored but trailing zeroes after decimal point are not ignored. So we count from 9 and last significant digit is 0.

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Location is 3 and bar is over 1. This is because it is not a number with decimal point. So the trailing zeroes are ignored. The count starts from 9 and last significant is 1.

7) 4.7 x 10⁻⁸

4.7 x 10⁻⁸

Location is 2 and bar is over 7. This is because the starting zeroes in a number with a decimal point are ignored. So the first digit considered is 4 and last significant digit is 7. If we expand this number:

4.7 x 10⁻⁸ =    0.000000047 = 0.000000047

Here the starting zeroes are ignored because there is a decimal point in the number.

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10,800,000.0

Location is 9 and bar is over 0.  Number has a decimal point so ending zero is not ignored and last significant figure is 0.

However if the number is like:

10,800,000. Then location would be 8 and bar is over 0.

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3.01 x 10²¹

Location is 3 and bar is over 1. Lets expand this number first

3.01 x 10²¹ = 3.01 x 1000000000000000000000

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So this is the number:

3010000000000000000000

Since this is number does not have a decimal point so the trailing zeroes are ignored. Hence the count starts from 3 and the last significant figure is 1

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0.000410

Location is 3 and bar is over 0. This is because the number has a decimal point so the ending zero is not ignored but the starting zeroes are ignored according to the rules given above. Hence the first significant figure is 4 and last significant figure is 0.

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