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yaroslaw [1]
3 years ago
5

A 20.0 mL sample of hydrochloric acid was neutralized with 58.4 mL of 0.250 M NaOH, in a titration.

Chemistry
1 answer:
erik [133]3 years ago
5 0

Answer: The Answer is 0.73.

Explanation: Solved in the attached picture.

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Howtocalculatethevolumeofcarbondioxideproducedwhen400gofmarblewereats.t.p<br>​
lys-0071 [83]

Answer:

so 0.15 moles X 22.4 dm3/mole=3.36 dm3. Next we find the moles of hexane combusted, and then the moles of CO2. Finally, we find the volume of CO2 using the fact that at STP, 1 mole of gas = 22.4 dm3.

3 0
3 years ago
Which scientific discipline belongs in the blue box?
Brums [2.3K]

Answer:

chemistry I think.

Explanation:

I don't need a thanks.

7 0
3 years ago
Read 2 more answers
Reema took 5ml of Lead Nitrate solution in a beaker and added approximately 4ml of Potassium Iodide solution to it. What would s
Brrunno [24]

She would observe a yellowish  solid precipitate which is the lead iodide and a white solid precipitate which is the potassium nitrate.

This is because the lead nitrate solution which contains particles of lead will mix with the potassium iodide solution containing  particles of iodide. Upon mixing,the  lead particles  from the Lead nitrate solution combines with the  iodide particles from Potassium iodide and form two compounds, a yellowish solid precipitate called lead iodide and a white solid  precipitate called Potassium nitrate.

The formation of entirely two new compounds is known as the double displacement reaction and can be written in a chemical equation as  

2KI(aq.)+Pb(NO₃)₂(aq.)------>2KNO₃(aq.)+PbI ₂(s)

See similar answer here  :https://brainly.in/question/46262462

4 0
2 years ago
Read 2 more answers
Assuming that the solubility of radon in water with 1 atm pressure of the gas over the water at 30 ∘C is 7.27×10−3M, what is the
Daniel [21]

Answer:

K = 137.55 atm/M.

Explanation:

  • The relationship between gas pressure and the  concentration of dissolved gas is given by  Henry’s law:

<em>P = (K)(C)</em>

where P is the partial pressure of the gaseous  solute above the solution (P = 1.0 atm).

k is a constant (Henry’s constant).

C is the concentration of the dissolved gas (C = 7.27 x 10⁻³ M).

∴ K = P/C = (1.0 atm)/(7.27 x 10⁻³ M) = 137.55 atm/M.

4 0
3 years ago
Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock sol
prohojiy [21]

Hey there!:

From the given data ;

Reaction  volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:

[E]t in molar  = g/L * mol/g

[E]t  = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

=758000 mole/L/min => 758000 M

Therefore :

Kcat = Vmax/ [E]t

Kcat = 758000 / 2.2*10⁻⁷ M

Kcat = 3.41441 *10¹² / min

Kcat = 3.41441*10¹² / 60 per sec

Kcat = 5.7*10¹⁰ s⁻¹

Hence   kcat of   xyzase is  5.7*10¹⁰ s⁻¹


Hope that helps!



4 0
3 years ago
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