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atroni [7]
3 years ago
10

What is chlorines density at room temperature (22°c)?​

Chemistry
1 answer:
Margaret [11]3 years ago
6 0

Answer:

Bromine (Br, element 35), also found as a diatomic molecule (Br2), is a liquid at room temperature, solidifying at -7.2ºC. GOOD LESSONS ♡

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What prefix should be used to name a hydrocarbon that has a six-carbon parent chain?a.hex- b.meth- c.non- d.sept-
liberstina [14]
A. hex- Normally with 6, the prefix is hex-. For example a polygon having 6 sides is known as hexagon.
8 0
4 years ago
As a chemist for an agricultural products company, you have just developed a new herbicide,"Herbigon," that you think has the po
Ganezh [65]

Answer:

pH = 2.03

Explanation:

The pH can be calculated using the following equation:

pH = -log [H_{3}O^{+}]  (1)

The concentration of H₃O⁺ is calculated using the dissociation constant of the next reaction:

CH₃COOH + H₂O ⇄  CH₃COO⁻ + H₃O⁺    

   1.00 M    

K_{a} = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}

Solving the above equation for H₃O⁺, we have:    

[H_{3}O^{+}] = \frac{Ka*[CH_{3}COOH]}{[CH_{3}COO^{-}]}    (2)    

The dissociation constant is equal to:    

pKa = -log(Ka) \rightarrow Ka = 10^{-pKa} = 10^{-4.76} = 1.74 \cdot 10^{-5}    

Now, by solving the equation of the solubility product for Herbigon, we can find [CH₃COO⁻]:

CH₃COOX  ⇄  CH₃COO⁻ +  X⁺  

                                             5.00x10⁻³ M

K_{sp} = [CH_{3}COO^{-}][X^{+}]

[CH_{3}COO^{-}] = \frac{K_{sp}}{[X^{+}]} = \frac{9.40 \cdot 10^{-6}}{5.00 \cdot 10^{-3}} = 1.88 \cdot 10^{-3} M

By entering the values of [CH₃COO⁻] and Ka, into equation (2) we can calculate [H₃O⁺]:

[H_{3}O^{+}] = \frac{1.74 \cdot 10^{-5}*[1.00]}{[1.88 \cdot 10^{-3}]} = 9.26 \cdot 10^{-3} M

Hence, the pH is:

pH = -log [H_{3}O^{+}] = -log [9.26 \cdot 10^{-3}] = 2.03

Therefore, the pH must be 2.03 to yield a solution in which the concentration of X⁺ is 5.00x10⁻³M.

I hope it helps you!  

6 0
4 years ago
(04.05 HC)
Tanzania [10]

Part 1;

The answer is; Carbon dioxide. Approximately 0.8% of the atmosphere is composed of carbon dioxide. It is in the gaseous phase.


Part 2;

The answer is; Combustion. Human activity of burning fossil fuels increases the carbon dioxide in the atmosphere. This has a secondary consequence of warming up the planet because carbon dioxide is a greenhouse gas.


Part 3:

Carbon is conserved in the cycles mostly in plants. Plants take up carbon dioxide in photosynthesis and make organic compounds (carbohydrates). Therefore first acts as carbon sinks. Even the fossils fuels that we burn come from plankton that is major carbon sinks too.


Part 4;

Carbon dioxide in the atmosphere is created by several processes including combustion and respiration. All living things respire out carbon dioxide. Plants take up this carbon dioxide and sequester it. When animals feed on plants they release part of this carbon back to the atmosphere through respiration


6 0
3 years ago
How many moles of water can be produced from the reaction of 28g of C3H8
Montano1993 [528]
2.55 moles H20 will be produced
7 0
3 years ago
A laser pulse with wavelength 525 nm contains 4.40 mj of energy. How many photons are in the laser pulse
Alexandra [31]

The laser pulse in this question has a wavelength of \lambda=524 nm=525\times 10^{-9}m. To solve this problem, we first have to calculate the energy of a single photon in the laser pulse. The equation for calculating the energy of a single photon of an electromagnetic wave is E=\frac{hc}{\lambda} where c is the speed of light, h is planks constant and \lambda is the wave length of the photons.

For this problem, c=3.0\times 10^8m/s, h=6.63\times10^{-34}J.s and \lambda=525\times 10^{-9}m. We use these values to calculate the energy of the photon as shown below,

E=\frac{hc}{\lambda} \\E=\frac{(6.63\times 10^{32}Js)\times(3.0\times10^8m/s)}{525\times 10^{-9}m} \\E=3.79\times 10 ^{-19}J.

Now that we know the energy for a single photon, we will divide the total energy given by the energy of one photon to get the number of photons in the pulse. The number of photons n is calculated as shown below,

n=\frac{4.4\times 10^{-3}J}{3.79\times10^{-19}J} =1.16\times 10^{16}. There are 1.16\times 10^{16} photons.

5 0
3 years ago
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