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olasank [31]
3 years ago
9

Why do you think some of the indicators used in experiment 2 were different than the ones used in experiment 1?

Chemistry
1 answer:
pentagon [3]3 years ago
6 0

During selection of indicator. We choose an indicator which have pH range equivalent to the pH change of reaction to give better result and better observation.

So there are some different indicator are used in table 2 as compared to the table 1. 

- Alizarin and phenolphthalein are basic indicator and their pH range is more than 8 so they are used in table 2

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3 years ago
A 0.885 M solution of KBr whose initial volume is 82.5 mL has more water added until its concentration is 0.500 M. What is the n
4vir4ik [10]

Answer:

V_2=146mL

Explanation:

Hello there!

In this case, since the equation for the calculation of dilutions is:

M_1V_1=M_2V_2

Whereas M is the molarity and V the volume, because the final concentration is lower than the initial. Thus, since we are asked to calculate the final volume, we solve for V2 as follows:

V_2=\frac{M_1V_1}{M_2}=\frac{0.885M*82.5mL}{0.500M}\\\\V_2=146mL

Best regards!

4 0
3 years ago
If 28 ml of 5.8 m h2so4 was spilled, what is the minimum mass of nahco3 that must be added to the spill to neutralize the acid?
ipn [44]

First we have to refer to the reaction between the acid and the base: <span>

H2SO4 + 2 NaHCO3 ---> 2 H2O + 2 CO2 + Na2SO4 

From this balanced equation we can see that for every 1 mol of acid (H2SO4), we need 2 mol of base (NaHCO3) to neutralize it. Given 28 ml of 5.8 M acid, we need to find out how many mols of acid that is: 

<span>28mL * (1L/1000mL) * 5.8 mol/L =  0.1624 mol H2SO4</span></span>

<span>
Since we need 2 mol of base per mol of acid, we need:</span>

<span> 2*0.1624 mol = 0.3248 mol NaHCO3 </span><span>

MolarMass of NaHCO3 is 84.01 g/mol 

<span>0.3248 mol*(84.01g/mol) = 27.29 g NaHCO3</span></span>

3 0
3 years ago
Read 2 more answers
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