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Troyanec [42]
3 years ago
8

Carry out the following calculation and report the answer using the proper number of significant figures: 38.251 + 73.1

Chemistry
2 answers:
Harrizon [31]3 years ago
8 0

Answer:

E is your answer dude ...........m

sergiy2304 [10]3 years ago
5 0

<u>Answer:</u>

<em>111.4</em>

<u>Explanation:</u>

1. Non zero digits are always significant for example 2356 has 4 significant figures

2. Zeros between Non zero digits are always significant 3007 has 4 significant figures

3. In a decimal trailing zeros are significant for example 0.0003 has 1 significant figures and 0.003000 has 4 significant figures.

Here we see all the options contain NON ZERO digits.

Hence the digits present in  all the numbers are significant and count them to find it

So the Answers are  

A) 3 significant figures

B) 4 significant figures

C) 4 significant figures

D) 5 significant figures

E) 6 significant figures

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3 years ago
Write a balanced half-reaction for the reduction of dichromate ion to chromium ion in acidic aqueous solution. Be sure to add ph
Serhud [2]

Answer:

The balanced half-reaction for the reduction of dichromate ion to chromium ion in acidic aqueous solution:

Cr_2O_7^{2-}(aq)+14H^+(aq)+6e^-\rightarrow 2Cr^{3+}(aq)+7H_2O(l)

Explanation:

The reaction :

Cr_2O_7^{2-}\rightarrow Cr^{3+}

Balancing of reaction in an acidic medium:

Step 1: Balance the atoms in the reaction:

Cr_2O_7^{2-}(aq)\rightarrow 2Cr^{3+}(aq)

Step 2: Balance oxygen atom by adding water on the side where no oxygen or less oxygen atom is present;

Cr_2O_7^{2-}(aq)\rightarrow 2Cr^{3+}(aq)+7H_2O(l)

Step 3: Balance the hydrogen atom by adding  hydrogen ions on the side where water is absent:

Cr_2O_7^{2-}(aq)+14H^+(aq)\rightarrow 2Cr^{3+}(aq)+7H_2O(l)

Step 4: Now balance charge by adding electrons on the side where more positive charge is present

Cr_2O_7^{2-}(aq)+14H^+(aq)+6e^-\rightarrow 2Cr^{3+}(aq)+7H_2O(l)

The balanced half-reaction for the reduction of dichromate ion to chromium ion in acidic aqueous solution:

Cr_2O_7^{2-}(aq)+14H^+(aq)+6e^-\rightarrow 2Cr^{3+}(aq)+7H_2O(l)

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Answer:

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