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goldfiish [28.3K]
3 years ago
13

What is the reducing agent in the reaction below?

Chemistry
2 answers:
frez [133]3 years ago
4 0
Option (a) is correct.
A reducing agent is the one which loses electrons to other substance and an oxidizing agent is one which accepts electrons. 
Here, In CrO_{3}, Cr has oxidation number 6+  in the L.H.S of the equation, but on R.H.S its oxidation number is 0 i.e. it Cr has gained electrons such that total charge is 0.
And the oxidation state of Al in the left-hand side of equation is 0 and in right-hand side, it is +6.i.e. it has donated its electrons to Cr. 

Hence, Cr is the oxidizing agent and Al is the reducing agent. 
igomit [66]3 years ago
3 0
Al is oxidized and Cr is reduced. So Al is the reducing agent. 

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A compound in a copper ore has the following percentage composition by mass:
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1) since we are given percentages, we can assume we have 100 grams of the molecule.

55.6 % Cu ----> 55.6 grams Cu
16.4 % Fe------> 16.4 grams Fe
28.0% S--------> 28.0 grams S

2) convert each gram to moles using the molar masses given

55.6 g Cu \frac{1 mol}{63.5 g} = 0.876 mol Cu
16.4 g Fe \frac{1 mol}{56.0 g} = 0.293 mol Fe
28.0 gS \frac{1 mol}{32.0} = 0.875 mol S

3) we divide the smallest value of moles (0.293) to each one.

Cu --> 0.876 / 0.293= 3
Fe---> 0.293 / 0.293= 1
S-----> 0.875 / 0.293= 3

4) let's write the empirical formula

Cu₃FeS₃
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At what temperature does water freeze?
Wewaii [24]

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Liquid A has a vapor pressure of 264 torr at 20∘C, and liquid B has a vapor pressure of 96.5 torr at the same temperature. If 5.
Katena32 [7]

Answer: 161.8 torr

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

p_1=x_1p_1^0 and p_2=x_2P_2^0

where, x = mole fraction

p^0 = pressure in the pure state

According to Dalton's law, the total pressure is the sum of individual pressures.

p_{total}=p_1+p_2p_{total}=x_Ap_A^0+x_BP_B^0

x_{A}=\frac{\text {moles of A}}{\text {moles of A+moles of B}}=\frac{5.50}{5.50+8.50}=0.39,

x_{B}=\frac{\text {moles of B}}{\text {moles of A+moles of B}}=\frac{8.50}{5.50+8.50}=0.61,

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p_{total}=0.39\times 264+0.61\times 96.5=161.8torr

The total vapor pressure above the solution is 161.8 torr.

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