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frutty [35]
3 years ago
12

What is the boiling point of 5 ml of alcohol

Chemistry
1 answer:
NeTakaya3 years ago
4 0
173.1f is the answer I believe, please let me know if I'm wrong then I would try to make up for it
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Can someone please help me with this, i'll give you a 5 star rating and the brainliest answer!​
elena-s [515]

Answer:

A. equation

B. atoms

C. products

Explanation:

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You wish to add 5 mg/l naocl as cl2 to a solution in a disinfection test, and you have a stock solution (household bleach) that
Alecsey [184]

Answer: -

0.1 ml of bleach should be added to each liter of test solution.

Explanation:-

Let the volume of bleach to be added is B ml.

Density of stock solution = 1.0 g/ml

Mass of stock solution = Volume of stock x density of stock

                                     = B ml x 1.0 g/ml

                                     = B g

Amount of NaOCl in this stock solution = 5% of B g

                                     = \frac{5}{100} x B g

                                     = 0.05 B g

Now each test solution must be added 5 mg/l NaOCl.

Thus each liter of test solution must have 5 mg.

Thus 0.05 B g = 5 mg

                        = 0.005 g

B = \frac{0.005}{0.05}

  = 0.1

Thus 0.1 ml of bleach should be added to each liter of test solution.

4 0
3 years ago
The chart below includes images of different structures found in animals or plants. For each structure, describe the smaller liv
Lady_Fox [76]
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3 years ago
Which produces the most energy? A. passing 5 kg of water through a hydroelectric power plant. B. fissioning 5 kg of uranium. C.
blagie [28]

Answer:

D. fusing 5 kg of hydrogen.

Explanation:

8 0
3 years ago
In a constant‑pressure calorimeter, 60.0 mL of 0.300 M Ba(OH)2 was added to 60.0 mL of 0.600 M HCl. The reaction caused the temp
Zepler [3.9K]

Answer:

ΔH = 57.04 Kj/mole H₂O

Explanation:

60ml(0.300M Ba(OH)₂(aq) + 60ml(0.600M HCl(aq)

=> 0.06(0.3)mole Ba(OH)₂(aq) + 0.60(0.6)mole HCl(aq)

=> 0.018mole Ba(OH)₂(aq) + 0.036mole HCl(aq)

=> 100% conversion of reactants => 0.018mole BaCl₂(aq) + 0.036mole H₂O(l) + Heat

ΔH = mcΔT/moles H₂O <==> Heat Transfer / mole H₂O

=(120g)(4.0184j/g°C)(27.74°C - 23.65°C)/(0.036mole H₂O)

ΔH = 57,042 j/mole H₂O = 57.04 Kj/mole H₂O

3 0
3 years ago
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