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Yuki888 [10]
3 years ago
15

Forrest spills some acid on the lab floor. He notifies his teacher and begins to look for baking soda to neturalize the acid on

the floor. Who should be notified next ?
Chemistry
2 answers:
Romashka [77]3 years ago
8 0

Explanation:

Since, acid is spilled on the lab floor so, first of all the teacher must be notified about it. As teacher can better tell you what should be done in such a case.

At the same time, rest of the students present in the lab must be notified about it so that no one can accidentally step onto the acid otherwise it might cause a serious injury to anyone.

Thus, we can conclude that rest of the students or classmates should be notified next in the given situation.

gregori [183]3 years ago
7 0
His classmates should be notified next
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The density of a pure substance is its mass per unit volume. The density of cresol has been measured to be 1024 g/L . Calculate
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Answer: The mass of 405 ml of cresol is 415 grams

Explanation:

Density is defined as the mass contained per unit volume.

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Putting in the values we get:

1024g/L=\frac{mass}{0.405L}

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Provide the reagents necessary to convert (S)-3-methyl-3-phenylpentanoic acid to (5)-3-methyl-3-phenylpentanamine.
8090 [49]

Answer:

1. SOCl2/pyridine

2. NH3

3. LiAlH4

4. H2O

Explanation:

In the conversion of  (S)-3-methyl-3-phenylpentanoic acid to (5)-3-methyl-3-phenylpentanamine, SOCl2/pyridine is first added to the  (S)-3-methyl-3-phenylpentanoic acid

This is then followed by the addition of NH3 is subsequently added followed by reduction using LiAlH4 which reduces the carbonyl carbon to an alkane. addition of water completes the mechanism and leads to the formation of the product.

3 0
3 years ago
A student used 10mL water instead of 30mL for extraction of salt from mixture. How may this change the percentage of NaCl extrac
34kurt
It will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.

If it is known that solubility of NaCl is 360 g/L, let's find out how many NaCl is in 30 mL of water:

360 g : 1 L = x g : 30 mL

Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = <span>x g : 30 mL

Now, crossing the products:
x </span>· 1,000 mL = 360 g · 30 mL
x · 1,000 mL = 10,800 g mL
x = 10,800 g ÷ 1,000 
x = 10.8 g

So, from 30 mL mixture, 10.8 g of NaCl could be extracted.

Let's calculate the same for 10 mL water instead of 30 mL.

360 g : 1 L = x g : 10 mL

Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = <span>x g : 10 mL

Now, crossing the products:
x </span>· 1,000 mL = 360 g · 10 mL
x · 1,000 mL = 3,600 g mL
x = 3,600 g ÷ 1,000 
<span>x = 3.6 g
</span>
<span>So, from 10 mL mixture, 3.6 g of NaCl could be extracted.
</span>
Now, let's compare:
If from 30 mL mixture, 10.8 g of NaCl could be extracted and <span>from 10 mL mixture, 3.6 g of NaCl could be extracted, the ratio is:
</span>3.6/10.8 = 1/3

Therefore, i<span>t will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.
</span>
7 0
3 years ago
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