Give an example to illustrate the difference between "paying more in taxes ” and paying a higher tax rate .

Convert 0.020 kg of Sn to mg of Sn

Lapatulllka [165]

hope that helps !

7 0

3 years ago

Pure nitrogen (N2) and pure hydrogen (H2) are fed to a mixer. The product stream has 40.0% mole nitrogen and the balance hydroge

LuckyWell [14K]

Explanation:

The given data is as follows.

Mass flow rate of mixture = 1368 kg/hr

$N_{2}$ in feed = 40 mole%

This means that $H_{2}$ in feed = (100 - 40)% = 60%

We assume that there are 100 total moles/hr of gas $(N_{2} + H_{2})$ in feed stream.

Hence, calculate the total mass flow rate as follows.

40 moles/hr of N_{2}/hr (28 g/mol of $N_{2}$ ) + 60 moles/hr of $H_{2}/hr$ (2 g/mol of $H_{2}$ )

$40 \times 28 g/hr + 60 \times 2 g/hr$

= 1120 g/hr + 120 g/hr

= 1240 g/hr

= $\frac{1240}{1000}$ (as 1 kg = 1000 g)

= 1.240 kg/hr

Now, we will calculate mol/hr in the actual feed stream as follows.

$\frac{100 mol/hr}{1.240 kg/hr} \times 1368 kg/hr$

= 110322.58 moles/hr

It is given that amount of nitrogen present in the feed stream is 40%. Hence, calculate the flow of $N_{2}$ into the reactor as follows.

$0.4 \times 110322.58 moles/hr$

= 44129.03 mol/hr

As 1 mole of nitrogen has 28 g/mol of mass or 0.028 kg.

Therefore, calculate the rate flow of $N_{2}$ into the reactor as follows.

$0.028 kg \times 44129.03 mol/hr$

= 1235.612 kg/hr

Thus, we can conclude that the the feed rate of pure nitrogen to the mixer is 1235.612 kg/hr.

3 0

3 years ago

Plan and create a model of complex carbohydrate that shows the relationship between simple and complex carbohydrates.

olga55 [171]

Answer:

Look at the image, It would help. Copy and paste if you need to

Explanation:

4 0

3 years ago

State 3 properties of acid

zaharov [31]

Explanation:

taste sour (vinegar)

smell frequently burns nose

texture sticky

5 0

3 years ago

List at least two chemical reactions that resulted in a white precipitate. For these reactions, list the possible chemical name

earnstyle [38]

Try looking through this, don't actually read it if you don't want to but if you skim through it you might find answers. Good Luck! :) ..Reactions of Aldehydes, Ketones And Phenols Objective: 1. To carry out some simple chemicals test in order to distinguish between aldehydes, ketones and phenols 2. To study the properties of aldehydes, ketones and phenols. 3. To identify the unknowns A, B, C, D and E. Introduction: Part I : Reaction of Aldehydes and Ketones The carbonyl group is C=O and any compound containing this group that can be described as a carbonyl compound. Carbonyl compounds fall into two main classes: aldehydes and ketones on the one hand and carboxylic acids and their derivatives on the other hand. The characteristic reactions of the aldehydes and ketones are addition and oxidation reactions occurring at the unsaturated carbonyl group. With the same reagent, aldehydes usually react faster than ketones, mainly because there is lees crowding at the carbonyl carbon and the steric effect. Aldehydes are also more easily oxidized than ketones. The carbonyl and other compounds investigated in this experiment are tested in each of the following ways: A) Chromic Acid (H2CrO4) Chromic acid is a strong oxidant. Aldehydes are oxidized to carboxylic acids by chromic acid. The Cr6+ in the chromic acid which is orange, then is reduced to Cr3+ which is green/blue. Ketones are not oxidized by chromic acid.image B) Tollen’s Test (Ag(NH3)2+ / OH- Tollen’s reagent (Ag(NH3)2+ / OH- is a weak oxidant. Aldehydes are readily oxidized to carboxylic acids by Tollen’s reagent to produce a silver mirror on the inside of a clean test tube. Ketones are not oxidized by Tollen’s reagent.image C) Fehling’s solution Fehling’s solution is an oxidizing agent. It is prepared by mixing equal part of Fehling’s solution I (copper(II) sulfate) and Fehling’s solution II (sodium potassium tartate and sodium hydroxide). Aldehydes are easily oxidized to carboxylic acid by Fehling’s solution and will reduce the cupric ion which complexed with tartate ion to cuprous oxide. A positive result is indicated by the formation of a brick red precipitate. Ketones are not oxidized by Fehling’s solution.image D) 2,4-dinitrophenylhydrazine (DNP Test or Brady’s Reagent) 2,4-dinitrophenylhydrazine (Brady’s reagent) is an important reagent related to hydrazine. Most aldehydes and ketones very readily with this reagent to give the yellow orange and red precipitates of 2,4-dinitrophenylhydrazones. Unconjugated aldehydes and ketones give precipitates toward the yellow while conjugated compound tend to be deeper colour of red. The conversion of aldehydes and ketones into hydrazone is an example of the addition-elimination reaction occurring at the unsaturated carbonyl group.image E) Iodoform Test Iodoform test can be used for the detection of acetalaldehyde and all methyl ketone which have the formula:image Iodoform, CHI3 is a yellow solid with a strong medicinal smell. Iodoform will precipitate out of a mixture of methyl ketone, iodine and base.image For acetaldehyde, the following reaction shows the formation of iodoform:image Compounds that are easily oxidized to acetaldehyde and methyl ketones also give a positive iodoform test. Only ethanol can be oxidized to acetaldehyde and secondary alcohol that have the general formula CH3CH(OH)R can be oxidized to methyl ketones. Part II: Reactions

5 0

3 years ago