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elena-14-01-66 [18.8K]
3 years ago
6

Prepare methoxyethane by Williamson's

Chemistry
1 answer:
8090 [49]3 years ago
8 0
If HI is present in excess and if temperature is high, methanol reacts with HI to form methyl iodide. (Not 100% sure)

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In the Minnesota Department of Health set a health risk limit for acetone in groundwater of 60.0 μg/L . Suppose an analytical ch
siniylev [52]

Answer:

m = 4.7 μg

Explanation:

Given data:

density of acetone = 60.0  μg/L

Volume = 79.0 mL

Mass = ?

Solution:

Formula:

d = m/v

v = 79.0 mL × 1L /1000 mL

v = 0.079 L

Now we will put the values on formula:

d = m/v

60.0  μg/L = m/0.079 L

m = 60.0 μg/L × 0.079 L

m = 4.7 μg

So health risk limit for acetone = 4.7  μg

3 0
3 years ago
What discovery led J.J. Thomson to decide that atoms were not indivisible, but are actually composed of smaller parts?
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Thomson used a beam of negatively charged particles. Using a beam of particles and detecting the scattering of the particles after they hit gold foil.
7 0
3 years ago
Organize the following species in terms
ser-zykov [4K]

Explanation:

  1. F Cl Br
  • ionization energy decreases we move down the group
  • so Br<Cl<F

2. Na K Li

  • ionization energy decreases we move down the group
  • so K<Na<Li

3. C N O F

  • ionization energy increases as we move across the peroid
  • so C<N<O<F

plz mark as brainliest if it helps

6 0
3 years ago
What two chemical elements make up the majority of our sun
Vsevolod [243]
The two  chemical elements that make up the <span>majority of our sun is :
"Hydrogen" and "Helium"

Hope this helps!</span>
8 0
3 years ago
Read 2 more answers
A laboratory instructor gives a sample of amino acid powder to each of four students, i, ii, iii, and iv, and they weigh the sam
Elena L [17]
Let us differentiate accuracy from precision. Accuracy is the nearness of the measured value to the true or exact value. On the other hand, precision is the nearness of the measured values between each other. So, for precision, select the student in which the measured values are very near to each other. That would be Student III. Now, for accuracy, let's find the average for each student.

Student I: (<span>8.72g+8.74g+8.70g)/3 = 8.72 g
Student II: (</span><span>8.56g+8.77g+8.83g)/3 = 8.72 g
Student III: (</span><span>8.50g+8.48g+8.51g)/3 = 8.50 g
Student IV: (</span><span>8.41g+8.72g+8.55g)/3 = 8.56 g

From the given results, the accurate one would be Students I and II. So, we make a compromise. Even though Student III is precise, it is not accurate. If you compare between Students I and II, the more precise data would be Student I. Therefore, the answer is Student I.</span>
6 0
3 years ago
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