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4 years ago

What is the name of this?

Chemistry

1 answer:

Ratling [72]4 years ago

7 0

Firstly, it’s all carbons with a double bond, making it an Alkene

The longest branch has 5 carbons, so we know that it’s a derivative of pentane.

We have a branch off of the 3rd carbon in the chain, no matter what end you start counting from.

CH2s are what we call Methylenes

So, it’s “3-methylenepentene”

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A pan containing 20.0 grams of water was allowed to cool from a temperature of 95.0 °C. If the amount of heat released is 1,200

Sedbober [7]

Answer:

81°C.

Explanation:

To solve this problem, we can use the relation:

Q = m.c.ΔT,

where, Q is the amount of heat released from water (Q = - 1200 J).

m is the mass of the water (m = 20.0 g).

c is the specific heat capacity of water (c of water = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).

∵ Q = m.c.ΔT

∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).

(- 1200 J) = 83.72 final T - 7953.

∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.

So, the right choice is: 81°C.

7 0

4 years ago

What is the pH of a 900 mL solution containing 3.40 grams of hydrocyanic acid

Allushta [10]

Answer:

$pH = 5.05$

Explanation:

pH is derived from the concentration of hydronium ions in a solution. Hydrocyanic acid is HCN.

First, we shall figure out the moles of HCN:

$\frac{3.4g}{27.03g/mol} = 0.125786$

If HCN was a strong acid:

HCN has a 1:1 ratio of H+ ions, the moles of H+ is also the same.

To find the molarity, we now divide by Liters. This gets us:

$\frac{0.125786 moles}{0.9L} = 0.139762 M$

Finally, we plug it into the definition of pH:

$pH = -log[H^{+} ]$

$pH = -log(0.139762)$

$pH = 0.855$

However, since HCN is a weak acid, it only partially dissociates. The $K_a$ of HCN is $6.2 * 10^{-10}$ .

$K_a = \frac{[H^+][A^-]}{[HA]}$

We can use an ice table to determine that when x = H+,

$K_a = \frac{x^2}{0.125786-x}$

$[H^+] = 8.83*10^{-6}$

$pH = -log[H^{+} ]$

$pH = -log(8.83 * 10^{-6} )$

$pH = 5.05$

7 0

4 years ago

A sample of helium gas occupies 355ml at 23°c. if the container the he is in is expanded to 1.50 l at constant pressure, what is

ss7ja [257]

Answer: The final temperature would be 1250.7 K.

Explanation: We are given a sample of helium gas, the initial conditions are:

$V_{initial}=355mL=0.355L$ (Conversion factor: 1L = 1000 mL)

$T_{initial}=23\°C=296K$ (Conversion Factor: 1° C = 273 K)

The same gas is expanded at constant pressure, so the final conditions are:

$V_{initial}=1.50L$

$T_{initial}=?K$

To calculate the final temperature, we use Charles law, which states that the volume of the gas is directly proportional to the temperature at constant pressure.

$V\propto T$