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kotykmax [81]
3 years ago
14

How would adding too much sodium hydroxide and overshooting the endpoint of the titration affect the concentration

Chemistry
1 answer:
Iteru [2.4K]3 years ago
4 0

Answer:

the concetration wouldent be equal and the ph would be off than the expected ph and it would have all ready passed the hydration point

Explanation:

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Which of the following ideas is true about matter? (Check all that apply.) *
elena-14-01-66 [18.8K]

Answer:

can be solid liquid or gas

is made of atoms and molecules

can change from one state to another without becoming a new substance just by adding or removing heat energy

Explanation:

5 0
3 years ago
Read 2 more answers
1.674×10-4 mol of an unidentified gaseous substance effuses through a tiny hole in 86.6 s. Under identical conditions, 1.715×10-
siniylev [52]
<h2>Answer:</h2>

44.06 g/mol

<h3>Explanation:</h3>

We are given;

  • Number of moles of unidentified gas as 1.674×10^-4 mol
  • Time of effusion of unidentified gas 86.6 s
  • Number of moles of Argon gas as 1.715×10^-4 mol
  • Time of effusion of Argon gas is 84.5 s

We are supposed to calculate the molar mass of unidentified gas

<h3>Step 1: Calculate the effusion rates of each gas</h3>

Effusion rate = Number of moles/time

Effusion rate of unidentified gas (R₁)

 =  1.674×10^-4 mol ÷ 86.6 s

 = 1.933 × 10^-6 mol/s

Effusion rate of Argon gas (R₂)

 = 1.715×10^-4 mol ÷ 84.5 sec

= 2.030 × 10^-6 mol/s

<h3>Step 2: Calculate the molar mass of unidentified gas</h3>
  • Assuming the molar mass of unidentified gas is x;
  • We can use the Graham's law of effusion to find x;
  • According to Graham's law of diffusion;

\frac{R_{1}}{R_{2}}}=\frac{\sqrt{MM_{Ar}}}{\sqrt{X}}

But, Molar mass of Argon is 39.948 g/mol

Therefore;

\frac{1.933*10^-6mol/s}{2.030*10^-6mol/s}}=\frac{\sqrt{39.948}}{\sqrt{X}}

0.9522=\frac{\sqrt{39.948}}{\sqrt{X}}

Solving for X

x = 44.06 g/mol

Therefore, the molar mass of the identified gas is 44.06 g/mol

3 0
4 years ago
A 150.0 mL sample of a 1.50 M solution of CuSO4 is mixed with a 150.0 mL sample of 3.00 M KOH in a coffee cup calorimeter. The t
svp [43]
Mols CuSO4 = M x L = 1.50 x 0.150 = 0.225 
<span>mols KOH = 3.00 x 0.150 = 0.450 </span>
<span>specific heat solns = specific heat H2O = 4.18 J/K*C </span>

<span>CuSO4 + 2KOH = Cu(OH)2 + 2H2O </span>
<span>q = mass solutions x specific heat solns x (Tfinal-Tinitial) + Ccal*deltat T </span>
<span>q = 300g x 4.18 x (31.3-25.2) + 24.2*(31.3-25.2) </span>
<span>dHrxn in J/mol= q/0.225 mol CuSO4 </span>
<span>Then convert to kJ/mol


</span>
5 0
4 years ago
Read 2 more answers
Help me please i need to sleep
lara [203]

Cloud is a gas.

ice is a solid.

snow is a solid.

steam is a gas.

rain is a liquid.

6 0
3 years ago
Dentify the reagents used to carry out the chlorination of benzene. a. cl2/ccl4 b. cl2/fecl3 c. cl2/nacl d. both b and c e. a, b
algol13

The reagents which are used to carry out the chlorination of benzene are   cl2/fecl3 and cl2/nacl

Chlorination is the process of disinfecting drinking water by introducing chlorine to eradicate viruses, germs, and parasites. To get drinking water with safe levels of chlorine, various techniques can be utilized. Chlorination is the technique of disinfecting and eradicating bacteria from drinking water by adding chlorine to it.

To get drinking water with safe levels of chlorine, various techniques can be utilized. Chlorine can be found as solid calcium hypochlorite (Ca(OCl)2), sodium hypochlorite solution (NaOCl), or compressed elemental gas. Chlorine is added to water as part of the chlorination process, however the chlorinating substance need not be pure chlorine.

To learn more about chlorination please visit -
brainly.com/question/14962130
#SPJ4

8 0
2 years ago
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