Another name for the acid test ratio is the

Please help me with this question!

asambeis [7]

I cant help u sorry .......

3 0

3 years ago

Four modes of sediment transport are:

netineya [11]

Here is 5

Dissolved Load - elements dissolved in solution

Suspended Load - very fine grained sediment such as clay and silt carried in suspension. The size grains that can be carried in suspension are dependent on the current velocity

Wash Load - a subset of the suspension load, extremely small particles (clay) that will remain in suspension independent of turbulence in the river

Saltation Load - particles that are temporarily carried in suspension but move by bouncing along the bottom

Bed Load - sediment that moves by rolling or sliding along the bottom. These are generally the coarser grained sediments such as sand and gravel.

5 0

3 years ago

A gaseous mixture composed of 20% CH4, 30% C2H4, 35% C2H2, and 15% C2H20. What is the average molecular weight of the mixture? a

Semenov [28]

Answer: The correct answer is Option d.

Explanation:

We are given:

Mass percentage of $CH_4$ = 20 %

So, mole fraction of $CH_4$ = 0.2

Mass percentage of $C_2H_4$ = 30 %

So, mole fraction of $C_2H_4$ = 0.3

Mass percentage of $C_2H_2$ = 35 %

So, mole fraction of $C_2H_2$ = 0.35

Mass percentage of $C_2H_2O$ = 15 %

So, mole fraction of $C_2H_2O$ = 0.15

We know that:

Molar mass of $CH_4$ = 16 g/mol

Molar mass of $C_2H_4$ = 28 g/mol

Molar mass of $C_2H_2$ = 26 g/mol

Molar mass of $C_2H_2O$ = 48 g/mol

To calculate the average molecular mass of the mixture, we use the equation:

$\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}$

where,

$\chi_i$ = mole fractions of i-th species

$m_i$ = molar masses of i-th species

$n_i$ = number of observations

Putting values in above equation:

$\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}$

$\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75$

Hence, the correct answer is Option d.

3 0

3 years ago

n acid with a pKa of 8.0 is present in a solution with a pH of 6.0. What is the ratio of the protonated to the deprotonated form

Nady [450]

Answer : The ratio of the protonated to the deprotonated form of the acid is, 100

Explanation : Given,

$pK_a=8.0$

pH = 6.0

To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}$

Now put all the given values in this expression, we get:

$6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}$

$\frac{[Deprotonated]}{[Protonated]}=0.01$

As per question, the ratio of the protonated to the deprotonated form of the acid will be:

$\frac{[Protonated]}{[Deprotonated]}=100$

Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100

5 0

3 years ago

Draw the structure(s) of the major organic product(s) of the following reaction. 1. lithium diisopropylamide / hexane 2. 1 eq. C

Jobisdone [24]

Answer:

See explanation below

Explanation:

You are not providing the starting material, however, I manage to find a similar question to this, so I'm gonna use it as a basis to help you answer yours.

Now let's analyze what is happening in the reaction so we can predict the final product.

We have a ketone here, reacting at first with LDA. This is a very strong base that is commonly used in reactions with ketones and aldehydes to promove a condensation. To do this, as LDA is a strong base it will occur firts an acid base reaction, substracting the most acidic hydrogen in the molecule (Which in this case, is the Beta hydrogen of the carbonile). This will cause an enolate formation.

Then, this enolate will react with the CH3I and form a new product. The final result would be a ketone with a methyl group now attached. In the picture 2, you have the mechanism and final product.

Hope this helps

6 0

3 years ago