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Elena L [17]
3 years ago
11

Camphor is reduced to isoborneol by sodium borohydride in ethanol. The instructions for safe disposal of the chemical waste gene

rated by this reaction state that the lids should not be screwed back onto the waste containers. Why should the lids not be put back on the waste containers?A) The reaction waste is still generating hydrogen gas that would get trapped in the container.B) Allowing the volatile waste to evaporate will lower the cost of chemical waste removal.C) Leaving the lid off allows the layers within the container to separate from one another.D) The tag on the container that lists its contents cannot be read properly when the lid is on.
Chemistry
2 answers:
LenKa [72]3 years ago
8 0

Answer:

The reaction waste is still generating hydrogen gas that would get trapped in the container.

Explanation: The chemical waste are dangerous and should be handled with care. The reactions don't stop immediately the products is formed putting the lid gives an enabling environment for chemical reactions to proceed.

Ronch [10]3 years ago
6 0

Answer:

A) The reaction waste is still generating hydrogen gas that would get trapped in the container.

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For a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k, calculate the inner diameter if you are designing for
bonufazy [111]

The inner diameter for a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k is 1.45 m

<h3>What is Stack Height ?</h3>

Stack height means the distance from the ground-level elevation at the base of the stack to the crown of the stack.

If a stack arises from a building or other structure, the ground-level elevation of that building or structure will be used as the base elevation of the stack.

Given is a steel stack that exhausts 1,200 cu.m/min of gases

P= 1 atm and

T= 400 K

maximum expected wind speed at stack height of 12 m/s

The formula for the diameter of chimney

\rm d=\sqrt{\dfrac{4Q}{\pi v} }

Q =1200 cu.m/min

= 1200 * 0.0166 = 19.92 cu.m/sec

Velocity = 12m/s

\rm d=\sqrt{\dfrac{4\times 19.92}{3.14*12} }\\

d= 1.45 m

Therefore The inner diameter for a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k is 1.45 m.

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8 0
2 years ago
12. In the modern periodic table, which of the following describes the elements with similar
V125BC [204]
B. same group I believe
8 0
3 years ago
Read 2 more answers
A solution is produced in which water is the solvent and there are four solutes. Which of the solutes can dissolve better if the
djverab [1.8K]

Answer:

What are the solutes

Explanation:

5 0
3 years ago
Nuclear decay<br>27Al + He 》 39P<br>​
dybincka [34]

Answer:

_{13}^{27}\text{Al} + \rm _{2}^{4}\text{He} \longrightarrow \, _{15}^{31}\text{P}

Explanation:

The unbalanced nuclear equation is

^{27}\text{Al} + \rm \text{He} \longrightarrow \, ^{31}\text{P}

We can insert the subscripts, because these are the atomic numbers of the elements

_{13}^{27}\text{Al} + \, \rm _{2}\text{He} \longrightarrow \, _{15}^{31}\text{P}

That leaves only the superscript of He to be determined,

The main point to remember in balancing nuclear equations is that the sums of the superscripts must be the same on each side of the equation.  

Then

27 + x = 31, so x = 31 - 27 = 4

Then, your nuclear equation becomes

_{13}^{27}\text{Al} + \, \rm _{2}^{4}\text{He} \longrightarrow \, _{15}^{31}\text{P}

6 0
3 years ago
The effusion rate of hcl is 43.2 cm/min in a certain effusion apparatus. what is the rate of effusion of ammonia in the same app
Jlenok [28]

The effusion rate is 1.125 cm/sec for ammonia.

How to find effusion rate ?

Effusion rate (r1) HCl = 43.2 cm/min

Molar mass (m2) NH3 =17.04g/mole

Molar mass (m1)  HCl    =36.46g/mole

  • Substitute the molar masses of the gases into Graham's law and solve for the ratio.
  • r1÷r2=√m2÷m1

       firstly convert 43.2 cm/min into cm/sec i.e., 0.72 cm/sec

      Then,

      0.72/r2 =√17.04/36.46

      r2= 1.125 cm/sec

Hence, the rate of diffusion of ammonia is 1.125 times faster than the rate of diffusion of hydrogen chloride.

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5 0
1 year ago
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