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snow_lady [41]
3 years ago
14

Which substance is oxidized and which is reduced in this reaction?

Chemistry
1 answer:
Eddi Din [679]3 years ago
6 0

Answer:

A) HBr oxidized; HCl reduced

Explanation:

Chemical equation:

Cl₂ + 2HBr → 2HCl + Br₂

The given reaction is redox reaction because oxidation reduction take place.

Cl₂ is reduced from oxidation state 0 to -1 while HBr oxidized from -1 to 0.

Chlorine is oxidizing agent while HBr is reducing agent.

Oxidation:

Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

Oxidizing agents:

Oxidizing agents oxidize the other elements and itself gets reduced.

Reducing agents:

Reducing agents reduced the other element are it self gets oxidized

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PLEASEEE HELPPPP I WILL AWARD BRAINLIEST
vampirchik [111]
Yesssirrrrrrrrr someone answer
3 0
3 years ago
Oxygen gas generated in the thermal decomposition of potassium chlorate is collected over water. The volume of gas collected is
Leya [2.2K]

The mass of oxygen collected from the thermal decomposition of potassium chlorate at a temperature of 297 K and 762 mmHg is 0.16 g

<h3>How to determine the mole of oxygen produced </h3>

We'll begin by obtaining the number of mole of oxygen gas produced from the reaction. This can be obtained by using the ideal gas equation as illustrated below:

  • Volume (V) = 0.128 L
  • Temperature (T) = 297 K
  • Pressure (P) = 762 – 22.4 = 739.6 mmHg
  • Gas constant (R) = 62.363 mmHg.L/Kmol
  • Number of mole (n) =?

PV = nRT

739.6 × 0.128 = n × 62.363 × 297

Divide both sides by 62.363 × 297

n = (739.6 × 0.128) / (62.363 × 297)

n = 0.0051 mole

Thus, the number of mole of oxygen gas produced is 0.0051 mole

<h3>How to determine the mass of oxygen collected</h3>

Haven obtain the number of mole of oxygen gas produced, we can determine the mass of the oxygen produced as follow:'

  • Mole = 0.0051 mole
  • Molar mass of oxygen gas = 32 g/mole
  • Mass of oxygen =?

Mole = mass / molar mass

0.0051 = mass of oxygen / 32

Cross multiply

Mass of oxygen = 0.0051 × 32

Mass of oxygen = 0.16 g

Thus, we can conclude that the mass of oxygen gas collected is 0.16 g

Learn more about ideal gas equation:

brainly.com/question/4147359

#SPJ1

5 0
1 year ago
If you have 400 grams of a substance that decays with a half-life of 14 days, then how much will you have after 56 days? To help
alisha [4.7K]

Answer:

25 grams

Explanation:

You strat off with 400 grams of your substance. By day 14, half has dcayed and you only have 200 grams left. By day 28, there are 100 grams of the substance. On day 42, there are 50 grams left. Finally, on day 56, the substance has been through four half-lives and 25 grams remain.

3 0
2 years ago
A solution has a [H3O+] of 1x 10-5 M. What is the [OH-] of the solution? (5 points)
White raven [17]

Answer:

1x10^–9 M

Explanation:

From the question given,

Concentration of hydronium ion, [H3O+] = 1x10^-5 M.

Concentration of Hydroxide ion, [OH-] =..?

The concentration of the hydroxide ion, [OH-] can be obtained as follow:

[H3O+] x [OH-] = 1x10^–14

1x10^-5 M x [OH-] = 1x10^–14

Divide both side by 1x10^-5

[OH-] = 1x10^–14 / 1x10^-5

[OH-] = 1x10^–9 M

8 0
3 years ago
The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p
saul85 [17]

Answer : The vapor pressure of bromine at 10.0^oC is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

T_2 = normal boiling point of bromine = 58.8^oC=273+58.8=331.8K

\Delta H_{vap} = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

Hence, the vapor pressure of bromine at 10.0^oC is 0.1448 atm.

4 0
3 years ago
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