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3 years ago

5,6-dimethyl-2-octyne

Chemistry

1 answer:

Natalija [7]3 years ago

5 0

See photo for drawing :)

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What is the main difference between active transport and passive transport? *

svp [43]

Answer:

There are two major ways that molecules can be moved across a membrane, and the distinction has to do with whether or not cell energy is used. Passive mechanisms like diffusion use no energy, while active transport requires energy to get done.

Explanation:

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3 years ago

For elements (Br, Ca, Fe, Na, S, Si, and Xe) which of the following term(s) apply?

ioda

Answer:

Detail is given below.

Explanation:

Bromine: Br (Halogen and non metal)

Bromine is present in group 17 and it is called halogen.

All halogens are very reactive.

Calcium; Ca ( alkaline earth metal)

Calcium is present in group 2. It is alkaline earth metal.

It has two valance electrons.

Iron; Fe (Transition metal)

Iron is transition metal and also called d-block element.

Sodium: Na (Metal, main group element)

Sodium is metal and present in group one.

It has one valance electron.

Sulfur: S (non metal)

Sulfur is non metal. It is present in group 16.

It has six valance electrons.

Silicon: Si (metalloid)

Silicon is metalloid. It is present in group 14.

It has four valance electrons.

Xenon: Xe (Noble gas, gas at a room temperature)

it is noble gas. Xenon is present in group 18.

It is gas at a room temperature.

3 0

3 years ago

Given their location on the periodic table, identify the ionic charge for each element and predict the compound formed by the ba

Maurinko [17]

Barium has a 2+ charge as it is in group 2 in the periodic table and so it has two electrons in its outer shell and chloride has a -1 charge on its chloride ion. So we will need two of the chloride ions as we have a 2+ charge to match the amount of charge on one barium ion- forming barium ion

BaCI2

8 0

3 years ago

Read 2 more answers

A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so

Xelga [282]

Answer : The equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

Explanation :

First we have to calculate the initial moles of $Fe^{3+}$ and $SCN^-$ .

$\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}$

$\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol$

and,

$\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}$

$\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol$

The given balanced chemical reaction is,

$Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)$

Since 1 mole of $Fe^{3+}$ reacts with 1 mole of $SCN^-$ to give 1 mole of $FeSCN^{2+}$

The limiting reagent is, $SCN^-$

So, the number of moles of $FeSCN^{2+}$ = 0.0020 mmole

Now we have to calculate the concentration of $FeSCN^{2+}$ .

$\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M$

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution

l = path length

$\epsilon$ = molar absorptivity coefficient

$\epsilon$ and l are same for stock solution and dilute solution. So,

$\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}$

For trial solution:

The equilibrium concentration of $SCN^-$ is,

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{initial}$ = 0.00050 M

Now calculate the $[FeSCN^{2+}]$ .

$C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M$

Now calculate the concentration of $SCN^-$ .

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)$

$[SCN^-]_{eqm}=4.58\times 10^{-8}M$

Therefore, the equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

5 0

4 years ago

What is the osmotic pressure of a solution made from 22.3 g of methanol (MM = 32.04 g/mol) that was added to water to make 321 m

xxMikexx [17]

Answer: The osmotic pressure of a solution is 53.05 atm

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = ?

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (methanol) = 22.3 g

Volume of solution = 321 mL

R = Gas constant = $0.0821\text{ L.atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$\pi=1\times \frac{22.3\times 1000}{32.04\times 321}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K$

$\pi=53.05atm$

Hence, the osmotic pressure of a solution is 53.05 atm

7 0

3 years ago