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o-na [289]
3 years ago
6

Which best describes the reducing agent in the reaction below? Cl2(aq) + 2Br(aq) 2Cl(aq) + Br2(aq) Bromine (Br) loses an electro

n, so it is the reducing agent. Bromine (Br) gains an electron, so it is the reducing agent. Chlorine (Cl) loses an electron, so it is the reducing agent. Chlorine (Cl) gains an electron, so it is the reducing agent.
Chemistry
2 answers:
MAVERICK [17]3 years ago
8 0

Answer:

A

Explanation:

Bromine (Br) loses an electron, so it is the reducing

SIZIF [17.4K]3 years ago
4 0

Answer:

Bromine (Br) loses an electron, so it is the reducing agent.

Explanation:

A reducing agent also called a reducer, is known to be an electron donor. A reducing agent is oxidized, because it loses electrons in the redox reaction.

A oxidising agent also called a oxidant or oxidiser, is known to be an electron acceptor. A oxidising agent is reduced, because it gains electrons in the redox reaction.

Cl2(aq) + 2Br-(aq) --> 2Cl-(aq) + Br2(aq)

Half ionic equations,

Cl2(aq) + 2e- --> 2Cl-(aq)

2Br-(aq) --> Br2(aq) + 2e-

Reducing agent = Br-

Oxidizing agent = Cl2

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Answer:

79

Explanation:

6 0
3 years ago
Which is an element?<br> A: H20<br> B: NaCl<br> C: Mg<br> D: CO
Talja [164]
The answer is option D "CO." Co also known as Cobalt is the 27th element on the periotic table. It was discovered in <span>1735, it's boiling point is 3200 k.</span>

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Hope this helps!

3 0
3 years ago
Read 2 more answers
The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) C
Kipish [7]

Answer: 72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Putting in the values we get:

\text{Number of moles}=\frac{19.7g}{44g/mol}=0.45moles

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)

According to stoichiometry:

1 mole of C_3H_8 requires 5 moles of oxygen

0.45 moles of C_3H_8 require= \frac{5}{1}\times 0.45=2.25 moles of oxygen

Mass of O_2=moles\times {\text {Molar mass}}=2.25\times 32=72g

72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

7 0
3 years ago
Identify the limiting reactant when 32. 0 g hydrogen is allowed to react with 16. 0 g oxygen
Mkey [24]

Answer:

Oxygen is limiting reactant

Explanation:

2 H2  +   O2  ======> 2 H2 O

from this equation (and periodic table) you can see that

  4 gm of H combine with 32 gm O2  

     H / O  =  4/32 = 1/8

       32 /16    =  2/1    shows O is limiter

         for 32 gm H you will need 256 gm O   and you only have 16 gm

3 0
2 years ago
An aqueous solution is 4.44 M nitric acid and the density of the solution is 1.42 g/mL. Calculate the mole fraction of this solu
prohojiy [21]

The mole fraction of HNO3 is  0.225

<u>Explanation:</u>

<u>1.</u>Given data

Density = 1.429 /ml

Mass% = 63.01 g HNO3 / 100g of solution

The mass of 63.01 g is in 100 / 1.142 /ml of solution

Or 63.01 g in 55.7 mL

Molarity = 15.39 moles / L

Mass of water in 100g = 100 - 63.01=36.99 g

So 63.01 grams in 36.99 grams of water

So mass of HNO3 in 1000grams of water = 63.01* x 1000 / 36.99 = 1703

Moles of HNO3 in 1000g = 1703 / 63.01 = 27.03 moles

Molality = 27.03 molal (mole / Kg)

Mole fraction = Mole of HN03 / Moles of water + mole of HNO3

Mole of water = 62/ 18 = 3.44

Moles of HNO3 = 63.01 / 63.01 = 1.000

Mole fraction = 1.000 / 3.44 + 1.000 = 0.225

The mole fraction of HNO3 is  0.225

7 0
3 years ago
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