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Zigmanuir [339]
3 years ago
11

Give an example of how a functional group can affect the properties of an organic compound.

Chemistry
1 answer:
Svetradugi [14.3K]3 years ago
7 0
The functional group can affect the chemical properties and physical properties. Like CH3CH3 and CH3CH2OH. The first one is gas and second one is liquid. The first one is not easy to react with others while the second one is.
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What’s 73m equal to in dm
MaRussiya [10]

Answer:730

Explanation:

6 0
4 years ago
What explains the fact that no machine is 100 percent efficient?
kati45 [8]
Hello, to answer your question. It is because all machines require a fuel or source to power it making it imperfect for using some source that may not be renewable.


                    Signed by, Virtouso Sargedog
3 0
3 years ago
The reaction of hydrogen and iodine to produce hydrogen iodide has a Kc of 54.3 at 703 K. Given the initial concentrations of H2
pentagon [3]

Answer:

[HI] = 0.7126 M

Explanation:

Step 1: Data given

Kc = 54.3

Temperature = 703 K

Initial concentration of H2 and I2 = 0.453 M

Step 2: the balanced equation

H2 + I2 ⇆ 2HI

Step 3: The initial concentration

[H2] = 0.453 M

[I2] = 0.453 M

[HI] = 0 M

Step 4: The concentration at equilibrium

[H2] = 0.453 - X

[I2] = 0.453 - X

[HI] = 2X

Step 5: Calculate Kc

Kc = [Hi]² / [H2][I2]

54.3 = 4x² / (0.453 - X(0.453-X)

X = 0.3563

[H2] = 0.453 - 0.3563 = 0.0967 M

[I2] = 0.453 - 0.3563 = 0.0967 M

[HI] = 2X = 2*0.3563 = 0.7126 M

3 0
3 years ago
What volume is needed to store 0.80 moles of helium gas at 204.6 kpa and 300 k?
shepuryov [24]
Answer is: 9,7 L is needed to store helium gas.
n(He) = 0,80 mol.
p(He) = 204,6 kPa.
T = 300 K.
R = 8,314 J/K·mol; universal gas constant.
Use ideal law eqaution: p·V = n·R·T.
V = n·R·T / p.
V(He) = 0,80 mol · 8,314 J/K·mol · 300 K ÷ 204,6 kPa.
V(He) = 9,75 L.


3 0
3 years ago
El agua del mar contiene aproximadamente un 3,0 % m/v de sal (NaCl, 58,44 g/mol), (asuma que es la única fuente de cloruros) si
Alchen [17]

Answer:

s = 4.41 g/L.

Explanation:

¡Hola!

En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:

PbCl_2(s)\rightleftharpoons 2Cl^-(aq)+Pb^{2+}(aq)

Lo cual hace que la expresión de equilibrio se calcule como:

Ksp=[Pb^{2+}][Cl^-]^2

Y que en términos de la solubilidad molar, s, se resuelve como:

1.6x10^{-5}=s(2s)^2\\\\1.6x10^{-5}=4s^3\\\\s=\sqrt[3]{\frac{1.6x10^{-5}}{4} } \\\\s=0.0159molPbCl_2/L

Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):

s=0.0159molPbCl_2/L*\frac{278.1gmolPbCl_2}{1molmolPbCl_2} \\\\s=4.41g/L

¡Saludos!

7 0
3 years ago
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