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2 years ago

_Cl2 + KI → KCI + __12

Chemistry

1 answer:

Snowcat [4.5K]2 years ago

8 0

Answer:

Cl2 +2Kl ----->2KCL +I2

Explanation:

_Cl2 + __KI → __KCI + __I2

Cl2 +2Kl ----->2KCL +I2

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FILL IN THE BLANK<br><br> an elements __________ is its row in the periodic table

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The question is asking to complete the said statement that is all about the element in the periodic table and base on my research and further investigation, I would say that the answer would be a Period. I hope you are satisfied with my answer and feel free to ask for more

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3 years ago

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Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0

klemol [59]

Answer:

Partial pressure of methane: 1.18 atm

Partial pressure of ethane: 1.45 atm

Partial pressure of propane: 2.35 atm

Explanation:

Let the total moles of gases in a container be n.

Total pressure of the gases in a container =P = 5.0 atm

Temperature of the gases in a container =T = 23°C = 296.15 K

Volume of the container = V = 10.0 L

$PV=nRT$ (Ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.0 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.0564 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16.04 g/mol}=0.4878 mol$

Moles of ethane gas = $n_2=\frac{18.00 g}{30.07 g/mol}=0.5986 mol$

Moles of propane gas = $n_3=?$

$n=n_1+n_2+n_3$

$n_3=n-n_1-n_2=2.0564 mol-0.4878 mol-0.5986 mol= 0.9700 mol$

Partial pressure of all the gases can be calculated by using Raoult's law:

$p_i=P\times \chi_i$

$p_i$ = partial pressure of 'i' component.

$\chi_1$ = mole fraction of 'i' component in mixture

P = total pressure of the mixture

Partial pressure of methane:

$p_1=P\times \chi_1=P\times \frac{n_1}{n_1+n+2+n_3}=P\times \frac{n_1}{n}$

$p_1=5.00 atm\times \frac{0.4878 mol}{2.0564 mol}=1.18 atm$

Partial pressure of ethane:

$p_2=P\times \chi_2=P\times \frac{n_2}{n_1+n+2+n_3}=P\times \frac{n_2}{n}$

$p_2=5.00 atm\times \frac{0.5986 mol}{2.0564 mol}=1.45 atm$

Partial pressure of propane:

$p_3=P\times \chi_3=P\times \frac{n_3}{n_1+n+2+n_3}=P\times \frac{n_3}{n}$

$p_3=5.00 atm\times \frac{0.9700 mol}{2.0564 mol}=2.35 atm$

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3 years ago

A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C.

Kay [80]

The equilibrium constant for this reaction at 350°C is D. 282.

<h3>Equilibrium constant</h3>

A dynamic chemical system approaches chemical equilibrium constant when enough time has passed and its composition no longer exhibits any discernible propensity to change further. The equilibrium constant of a chemical reaction is the value of its reaction quotient in this condition. The equilibrium constant is independent of the initial analytical concentrations of the reactant and product species in the mixture for a specific set of reaction conditions. Understanding equilibrium constants is crucial for comprehending many chemical systems as well as biological processes like the transport of oxygen by hemoglobin in the blood and the maintenance of acid-base homeostasis in the human body. There are many different kinds of equilibrium constants, including stability constants, formation constants, binding constants, association constants, and dissociation constants.

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A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C.

Br2(g) + I2(g) ↔ 2IBr(g)

When the mixture has come to equilibrium, the concentration of iodine monobromide is

1.190 M. What is the equilibrium constant for this reaction at 350°C? Show step-by step explanation.

A) 3.55 × 10^3

B) 1.24

C) 1.47

D) 282

E) 325

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1 year ago

What is the unit of measurement used to express density

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Density Grams or kilograms
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If you wanted to find a sample of fermium, which has an atomic number 100, where would you look?

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The answer would be B: In the Earth’s atmosphere!.

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2 years ago

_Cl2 + __KI → __KCI + __12

_Cl2 + KI → KCI + __12