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Vikki [24]
2 years ago
13

Function of starch solution​

Chemistry
2 answers:
serious [3.7K]2 years ago
5 0

Answer:

Starch is a viable indicator in the titration process because it turns deep dark blue when iodine is present in a solution. When starch is heated in water, decomposition occurs and beta-amylose is produced

Amiraneli [1.4K]2 years ago
4 0

In the titration process, starch is a viable indicator because when iodine is present in a solution, it turns deep dark blue. Decomposition happens when starch is heated in water, and beta-amylose is formed.
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MnO2 + 4HCl reacts to form MnCl2 + CL2 + 2H2O If .86 mol of MnO2 and 48.2 g of HCl react, which reactant is the limiting reagent
Katen [24]
HCl:
<span>
m=48,2g
M=36,5g/mol

n = m/M = 48,2g / 36,5g/mol = 1,32mol

1mol          : 4mol
MnO</span>₂        + 4HCl ⇒ MnCl₂ + Cl₂ + 2H₂O
0,86mol     :  1,32mol
                     limiting reagent 
0,33 will react

HCl is limiting reagent.
6 0
3 years ago
Elements in the periodic are listed according to
sammy [17]

Answer:

the raising atomic number

Explanation:

Elements are listed on the periodic table according to their atomic number.

6 0
3 years ago
You are using a ruler to measure the width of a test tube. The edge of the test tube is between 3 cm and 4 cm marks. if centimet
Arada [10]

Answer:

A) 3.6 cm

Explanation:

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So, the answer should be 3.6 cm.

Here's a document that explains it well: https://www.auburn.wednet.edu/cms/lib03/WA01001938/Centricity/Domain/1360/1_Uncertainty.pdf

Hope that's right!

5 0
1 year ago
If a person did overshoot the endpoint, would this cause the experimental value for the % of acid in vinegar to be too high or t
Stels [109]
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7 0
3 years ago
An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its
german
Hello!

First you need to calculate q 
<span>delta U is change in internal energy </span>

<span>delta U = q + w </span>
<span>q is heat and w work done </span>
<span>here work was done by the system means energy leaving the system so w is negative </span>

<span>delta U = q + w </span>

<span>q = delta U - w = 6865 J - (-346 J) = 7211 J = 7.211 KJ </span>

<span>q = m x c x delta T </span>

<span>7211 J = 80.0 g x c x (225-25) °C </span>

<span>c = 0.451 J /g °C
</span>
Hope this Helps! Have A Wonderful Day! :)
5 0
3 years ago
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