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777dan777 [17]
3 years ago
6

What element is being oxidized Fe2O3 + 3CO -> 2FE +3C02

Chemistry
1 answer:
Vinvika [58]3 years ago
7 0
Fe because oxidation mean loss of oxygen and Fe lose the oxygen so Fe is oxidised
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Four flasks are prepared, each containing 100mL of aqueous solutions of equal concentration. Which solution has the lowest freez
Jet001 [13]
The freezing point depression is a colligative property which means that it is proportional to the number of particles dissolved.

The number of particles dissolved depends on the dissociation constant of the solutes, when theyt are ionic substances.

If you have equal concentrations of two solutions on of which is of a ionic compound and the other not, then the ionic soluton will contain more particles (ions) and so its freezing point will decrease more (will be lower at end).   

In this way you can compare the freezing points of solutions of KCl, Ch3OH, Ba(OH)2, and CH3COOH, which have the same concentration.

As I explained the solution that produces more ions will exhibit the greates depression of the freezing point, leading to the lowest freezing point.

In this case, Ba(OH)2 will produce 3 iones, while KCl will produce 2, CH3OH will not dissociate into ions, and CH3COOH will have a low dissociation constant.

Answer: Then, you can predict that Ba(OH)2 solution has the lowest freezing point.
6 0
3 years ago
Help me asap Like fr i need to turn this whole edjenuity section by 2 40
lana [24]

Answer:

3/6 or 1/2

Explanation:

brainliest?

8 0
3 years ago
Ammonia, nh3, is a weak base with a kb = 1.8 x 10-5. in a 0.8m solution of ammonia, which has a higher concentration: ammonia (n
insens350 [35]
Answer is: concentration ammonia is higher than concentration of ammonium ion.
Chemical reaction of ammonia in water: NH₃ + H₂O → NH₄⁺ + OH⁻.
Kb(NH₃) = 1,8·10⁻⁵.
c₀(NH₃) = 0,8 mol/L.
c(NH₄⁺) = c(OH⁻) = x.
c(NH₃) = 0,8 mol/L - x.
Kb = c(NH₄⁺) · c(OH⁻) / c(NH₃).
0,000018 = x² /  0,8 mol/L - x.
solve quadratic equation: x = c(NH₄⁺) = 3,79·10⁻³ mol/L.
7 0
3 years ago
The rate at which a certain Australian tree cricket chirps is 194/min at 28°C, but only 47.6/min at 5°C, From these data calcula
uysha [10]

Answer: The energy of activation for the chirping process is 283.911 kJ/mol

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

The expression used with catalyst and without catalyst is,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_2 = rate of reaction at 28^0C = 194/min

K_1 = rate of reaction at  5^0C = 47.6 /min

Ea = activation energy

R = gas constant = 8.314 J/Kmol

tex]T_1[/tex] = initial temperature = 5^oC=273+5=278K

tex]T_1[/tex] = final temperature = 28^oC=273+28=301K

Now put all the given values in this formula, we get

\frac{194}{47.6}=\frac{E_a}{2.303\times 8.314}[\frac{1}{278}-\frac{1}{301}]

{E_a}=283911J/mol=283.911kJ/mol

Thus the energy of activation for the chirping process is 283.911 kJ/mol

8 0
4 years ago
What is the formula for heat of fusion?
spin [16.1K]

Answer:

The formula used to calculate heat of fusion:  

q = m·ΔH f

Explanation:

The formula used to calculate heat of fusion:  

q = m·ΔH f

8 0
3 years ago
Read 2 more answers
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