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Phantasy [73]
2 years ago
9

What three obstacles did dmitri mendeleev face in his life

Chemistry
2 answers:
Mama L [17]2 years ago
8 0
<span>~He was working without prior knowledge toward creating a periodic table of elements that had been done in Europe in the 1860s

~ scandal when he got married to his second wife before he divorced his first wife

~he was working with data that was not very precise (technology didn't exist at the time to make accurate measurements (e.g., weights) for the things he was working on.) Many elements had also not been discovered yet, so he had to guess at a lot of things when making his conclusions. </span>
OLga [1]2 years ago
6 0

Answer:

Answered

Explanation:

1. He married his second wife before giving divorce to his first wife. This made him an outsider in the scientific community making it difficult for him to raise funds for the research.

2. Back then there were no precise measuring instruments, so it was difficult for him to measure correct data. Many of the elements were yet to be discovered, so he guessed and by fortune most of his guesses were right.

3.The elements that were not discovered at that time, he left blank spaces for them.  

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HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.
Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, C_A = 6.0mol/dm³

volume of HA, V_A  = 25.0cm³, = 0.025dm³

Concentration of HB, C_B = 3.0mol/dm³

volume of HB, V_B = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

C_iVi = C_fV_f

where;

C_i is initial concentration

V_i is initial volume

C_f is final concentration of the solution

V_f is final volume of the solution

C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

7 0
2 years ago
Which of the following statements are True about the experimental process used in the Diels Alder reaction?
vovikov84 [41]

Here we have to get the correct statements among the given, applicable for Diels-Alder reaction.

The true statements in case Diels-Alder reaction are-

1. An excess of Maleic anhydride is used.

2. The I.R. of the products are indistinguishable.

The Diels-Alder reaction is the most is the most important cyclo-addition reaction in organic chemistry. These are addition reactions in which ring systems are formed without eliminating any compounds.

There remains one diene and one dienophile. The reaction is reversible in nature and requires elevated temperature to obtain its transition state. The reaction rate become faster in certain condition like using of polar solvents.

Among the given statements the following statements are true-

1. An excess of maleic anhydride (the most effective di-enophile) is used to process the reaction in forward direction.

2. The products obtain in this reaction are stereoisomers thus are indistinguishable by infrared spectroscopy (IR).

The statements which are not true for the Diels-Alder reaction:

3. The re-crystallization of the products by any polar solvent like methanol is not feasible as it will cause the retro reaction due to stability of the transition state in polar solvent.

4. Cleaning of glassware are compulsory for any reaction it is not specifically true for Diels-Alder reaction.

5. The reaction occurs at elevated temperature thus flame is required.

 

8 0
2 years ago
What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
Mashutka [201]

Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

c=1.4\times 10^{-4}

6 0
2 years ago
George splits a bagel in half and places both sides of the bagel in a toaster. He turns on the toaster, and within a minute the
STALIN [3.7K]
C. Convention

Explanation: I'm pretty sure it is

6 0
2 years ago
Read 2 more answers
84. Heavy water, D2O (molar mass = 20.03 g mol–1), can be separated from ordinary water, H2O (molar mass = 18.01), as a result o
ipn [44]

Answer:

relative rate of diffusion is 1.05

Explanation:

According to Graham's law of difussion:

Rate of diffusion is inversely proportional to the square root of molecular weight of a molecule.

For two given molecules:

\frac{(rate)_{1}}{(rate)_{2}}=\frac{\sqrt{M_{2}} }{\sqrt{M_{1}}}

The given molecules are

Water = 18.01

Heavy water =20.03

Thus the relative rate of diffusion will be:

\frac{(rate)_{water}}{(rate)_{heavywater}}=\sqrt{ \frac{20.03}{18.01}}=1.05

8 0
2 years ago
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