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3 years ago

What commercially available compound is used to generate cl2?

Chemistry

2 answers:

Kipish [7]3 years ago

7 0

Hello!

The commercially available compound that is used to generate Cl₂ is Sodium Hypochlorite

Sodium Hypochlorite is present in commercial bleach. Its chemical formula is NaClO.

Sodium Hypochlorite can react with acids (which can be present in vinegar, glass and window cleaners, drain cleaners or lime, calcium and rust removal products) to form Chlorine gas as shown by the following reaction (for a generic acid HA)

NaClO(aq) + HA(aq) → H₂O(aq) + Cl₂(g) + NaA(aq)

Chlorine gas is very toxic, so bleach should never be mixed with anything but water.

Have a nice day!

GuDViN [60]3 years ago

6 0

<span>Bleach and ammonia can be used to make cl2. Bleach is commercially available and so people should be very careful during these experiments or even when mixing things in the household.</span>

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Answer:

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2 years ago

A sample of O2 gas (2.0 mmol) effused through a pinhole in 5.0 s. It will take __________ s for the same amount of CO2 to effuse

ludmilkaskok [199]

Answer:

$\large \boxed{\text{5.9 s}}$

Explanation:

Graham’s Law applies to the effusion of gases:

The rate of effusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

$r \propto \dfrac{1}{\sqrt{M}}$

If you have two gases, the ratio of their rates of effusion is

$\dfrac{r_{2}}{r_{1}} = \sqrt{\dfrac{M_{1}}{M_{2}}}$

The time for diffusion is inversely proportional to the rate.

$\dfrac{t_{2}}{t_{1}} = \sqrt{\dfrac{M_{2}}{M_{1}}}$

Let CO₂ be Gas 1 and O₂ be Gas 2

Data:

M₁ = 44.01

M₂ = 32.00

Calculation

$\begin{array}{rcl}\dfrac{t_{2}}{t_{1}} & = & \sqrt{\dfrac{M_{2}}{M_{1}}}\\\\\dfrac{t_{2}}{\text{5 s}}& = & \sqrt{\dfrac{44.01}{32.00}}\\\\& = & \sqrt{1.375}\\t_{2}& = & \text{5 s}\times 1.173\\& = & \mathbf{5.9 s} \\\end{array}\\\text{It will take $\large \boxed{\textbf{5.9 s}}$ for the carbon dioxide to effuse.}$

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3 years ago

The combustion reaction for methane is shown below:

horrorfan [7]

Answer:

P = 14.1 atm

Explanation:

Given data:

Mass of methane = 64 g

pressure exerted by water vapors = ?

Volume of engine = 24.0 L

Temperature = 515 K

Solution:

Chemical equation:

CH₄ + 2O₂ → CO₂ + 2H₂O + energy

Number of moles of methane:

Number of moles = mass / molar mass

number of moles = 64 g/ 16 g/mol

Number of moles = 4 mol

Now we will compare the moles of water vapors and methane.

CH₄ : H₂O

1 : 2

4 : 2/1×4 = 8 mol

Pressure of water vapors:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

P = 8 mol × 0.0821 atm.L/mol.K× 515 K / 24.0 L

P = 338.25 atm.L/ / 24.0 L

P = 14.1 atm

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