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lapo4ka [179]
2 years ago
5

Draw the structures of the starting materials used to synthesize 3-methylbutyl propanoate. (draw the structures in the single sk

etchpad shown. draw only the structures - do not include a plus sign.)

Chemistry
1 answer:
Feliz [49]2 years ago
7 0
<span>3-Methylbutyl propanoate can be synthesized by reacting Propionic Acid with 3-Methylbutanol in the presence of Acid. This reaction is called as Esterification. The synthesis of </span><span>3-Methylbutyl propanoate is shown below,</span>

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If one has a solution of 0.10 m silver nitrate and it is diluted by a factor of two, what is the new concentration
Angelina_Jolie [31]
Diluted by a factor of two means that we double the volume of the solution by adding an equal volume of the water.
if we diluted it by a factor of one so the new concentration = 0.1/2=0.05 M and diluted by a factor of two so, the new concentration will be 0.05/2 = 0.025 M
7 0
2 years ago
Given that the density of TlCl(s) is 7.00 g/cm3 and that the length of an edge of a unit cell is 385 pm, determine how many form
ryzh [129]
Here's my best guess
the volume of the unit cell is (385*10^-12)^3=5.7066*10^-29 m^3
multiply by density to get mass mass = (7 g/cm^3)*(100^3 cm^3 / 1^3 m^3) * 5.7066*10^-29 m^3= 3.99466*10^-22 g
covert to moles 3.99466*10^-22 g * 1 mol / 239.82 g = 1.6657 *10^-24 mol
convert to number of units

1.6657 *10^-24 mol * 6.23*10^23 units/mol = 1.04

385 pm = 3.85*10^(-8) cm

The volume of the unit cell is the cube of that, which is 5.71*10^(-23) cm^3. Since the ratio of mass to volume (i.e. the density) must be the same no matter what amount of TlCl you have, you can say:
7 = x/(5.71*10^(-23)), where x is the mass of the unit cell. Solving for x, you get 4*10^(-22) g.

The mass of a molecule of TlCl is 240 amu, which in grams is 4*10^(-22) g. The mass of the unit cell and the mass of a molecule of TlCl is the same. Therefore there is one formula unit of TlCl per unit cell.

3 0
2 years ago
What is the pH of a 0.75 M HNO3 solution
Sauron [17]
Hello!

The dissociation reaction of HNO₃ is the following:

HNO₃ → H⁺ + NO₃⁻

This is a strong acid, so the concentration of HNO₃ would be the same as the concentration of H⁺. The formula for pH is the following:

pH=-log([H_3O^{+}])=-log(0,75M)=0,12

So, the pH would be 0,12

Have a nice day!
4 0
2 years ago
Read 2 more answers
Algunas drogas bloquean los químicos secretados por los axones. ¿Cómo podrían afectar estas drogas al impulso nervioso? ¿Qué pod
olya-2409 [2.1K]

Answer:

Algunas drogas intervienen en el impulso nervioso inhibiendo la liberacion de  neurotransmisores, por ende estas neuronas no generaran efectos postsinapticos.

Podria pasar que algunos procesos internos estaran inhibidos o fomentando indirectamente la aparicion de otros, un ejemplo claro de esto es la xerestomia generada por las drogas betabloqueantes.

Explanation:

Es asi que los procesos internos se veran desregularizados, las drogas que inhiben la liberacion de los neurotransmisores son reguladores por lo general de la tension arterial o antiarritmicos cardiacos, es por eso que su consumo es fundamental y vale el riesgo beneficio a la hora de evaluar la inhibicion de las respuestas postsinapticas.

Los neurotransmisores mas conocidos son la noradrenalina y adrenalia, le membrana presinaptica es la nerviosa y la postsinaptica puede ser muscularo o nerviosa.

4 0
2 years ago
3. A 31.2-g piece of silver (s = 0.237 J/(g · °C)), initially at 277.2°C, is added to 185.8 g of a liquid, initially at 24.4°C,
VARVARA [1.3K]

Answer:

Cp_{liquid}=2.54\frac{J}{g\°C}

Explanation:

Hello,

In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

Q_{Ag}=-Q_{liquid}

That in terms of the heat capacities, masses and temperature changes turns out:

m_{Ag}Cp_{Ag}(T_2-T_{Ag})=-m_{liquid}Cp_{liquid}(T_2-T_{liquid})

Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}

Best regards.

6 0
2 years ago
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