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masya89 [10]
2 years ago
13

All of the orbitals in a given subshell have the same value as the ________ quantum number.

Chemistry
1 answer:
Nezavi [6.7K]2 years ago
7 0
All of the orbitals in a given sub shell have the same value as the principal quantum number, I believe.
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What is the final product of the following sequence of reactions? (CH3NCHOH PBr3 Mg diethyl ether 1) L N PCC 2) H3O * CH2Cl2 A)
lubasha [3.4K]

Answer:

(CH3)2CHCH2CHO

Explanation:

The reaction sequence begins with the reaction of isopropanol with phosphorus tribromide to yield isopropyl bromide. This is followed by reaction with Magnesium in ether solution giving a grignard reagent, isopropyl magnesium bromide. This is now attacked by oxirane and the epoxide ring opens, hydrolysis of the product, followed by oxidation using pyridinium chlorochromate (PCC) yields the final product- (CH3)2CHCH2CHO

The detailed reaction mechanism is attached to this answer.

4 0
2 years ago
What is the mass in grams of one mole of many of any substance known as
OLEGan [10]
The molecular weight of the substance.
7 0
3 years ago
How much does one mole of NH3 weigh?
Leokris [45]
17.031 g/mol Hope this helps you
6 0
2 years ago
When a sample of aqueous hydrochloric acid was neutralized with aqueous sodium hydroxide in a calorimeter, the temperature of 10
Annette [7]
<span>6.50x10^3 calories. Now we have 4 pieces of data and want a single result. The data is: Mass: 100.0 g Starting temperature: 25.0°C Ending temperature: 31.5°C Specific heat: 1.00 cal/(g*°C) And we want a result with the unit "cal". Now you need to figure out what set of math operations will give you the desired result. Turns out this is quite simple. First, you need to remember that you can only add or subtract things that have the same units. You may multiply or divide data items with different units and the units can combine or cancel each other. So let's solve this: Let's start with specific heat with the unit "cal/(g*°C)". The cal is what we want, but we'ld like to get rid of the "/(g*°C)" part. So let's multiply by the mass: 1.00 cal/(g*°C) * 100.0 g = 100.0 cal/°C We now have a simpler unit of "cal/°C", so we're getting closer. Just need to cancel out the "/°C" part, which we can do with a multiplication. But we have 2 pieces of data using "°C". We can't multiply both of them, that would give us "cal*°C" which we don't want. But we need to use both pieces. And since we're interested in the temperature change, let's subtract them. So 31.5°C - 25.0°C = 6.5°C So we have a 6.5°C change in temperature. Now let's multiply: 6.5°C * 100.0 cal/°C = 6500.0 cal Since we only have 3 significant digits in our least precise piece of data, we need to round the result to 3 significant figures. 6500 only has 2 significant digits, and 6500. has 4. But we can use scientific notation to express the result as 6.50x10^3 which has the desired 3 digits of significance. So the result is 6.50x10^3 calories. Just remember to pay attention to the units in the data you have. They will pretty much tell you exactly what to add, subtract, multiply, or divide.</span>
4 0
3 years ago
a mixture is made up of 3.35g of SiO2, 0.38g of cellulose and 8.76g of calcium carbonate. What is the percent of cellulose in th
Alchen [17]

Solution:

mass of the cellulose in the mixture is 0.38g

total mass of the mixture is:

3.35+0.38+8.76

=12.4g

thus the  percentage of the cellulose in the mixture is:

mass of the cellulose/total mass of the mixture*100%

0.38/12.4*100%

=3%

6 0
3 years ago
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