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julia-pushkina [17]
4 years ago
5

Can manganese be a catalyst

Chemistry
1 answer:
Genrish500 [490]4 years ago
7 0
Manganese alone cannot be a catalyst. However, its oxides can work as a catalyst. Manganes (II, III) oxide has found some applications in certain reactions as a catalyst. These reactions are the oxidation of methane, carbon monoxide, decomposition of NO and the catalytic combustion of organic compounds.
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True or False: a mixture that has a uniform composition and uniform properties in all its parts is called a homogeneous mixture
sattari [20]

Answer:

True

Explanation:

The composition and properties are so uniform that it's difficult to physically differentiate them.

6 0
3 years ago
Which statement describes an intensive property of matter? It is the same for every sample o single substance. O It depends on h
Korolek [52]

Answer:

dpends on how the substance was formed. Just took the test.

Explanation:

3 0
4 years ago
Read 2 more answers
It takes 500 kJ to remove one mole of electrons from the atoms at the surface of a solid metal. How much energy does it take to
prisoha [69]

Answer:

8.31 × 10⁻²² kJ

Explanation:

Step 1: Given data

Energy required to remove one mole of electrons from the atoms at the surface of a solid metal: 500 kJ/mol e⁻

Step 2: Calculate how much energy does it take to remove a single electron from an atom at the surface of this solid metal

We will use Avogadro's number: there are 6.02 × 10²³ electrons in 1 mole of electrons.

500 kJ/mol e⁻ × 1 mol e⁻/6.02 × 10²³ e⁻ = 8.31 × 10⁻²² kJ/e⁻

7 0
3 years ago
Be sure to answer all parts. The rate law for 2 NO(g) + O2(g) → 2 NO2(g) is rate = k[NO]2[O2]. The following mechanisms have bee
rodikova [14]

Answer:

a. I

b. I

Explanation:

A rate law is an equation that relates the rate of a reaction to the concentration of reactants (and catalyst) raised to various powers.

In the equation above

2 NO(g) + O2(g) → 2 NO2(g)

rate = k[NO]²[O2]

where k = rate constant

From the mechanism above we see that in;

1: Rate law= k[NO]² [O2]

2: Rate law= k[N2O2][O2] [slow eq determines rate law]

3: Rate law= k[N2][O2]²

We can observe that the resembling equation is 1.

The rate of a chemical reaction is determined by the slowest step.

Rate = k[Concentration of reactants individually raised to their stoichiometric co-efficients]

In mechanism I,

Overall reaction occurs in a single step. Therefore,

rate= k[NO]²[O2]

This is this consistent with the observed rate law.

In mechanism II,

The overall reaction occurs in two steps, through the involvement of an intermediate, N2O2.

Rate of the slowest step should be the overall reaction rate.

Therefore, overall rate= k[N2O2][O2]

Again considering non-accumulation of intermediate, N2O2 in the overall reaction.

Its rate of production will be equal to its rate of decomposition.

Thus, k1[NO]²= k[N2O2][O2]

➡ [N2O2]= (k1/k). [NO]²/[O2]

Overall rate= k(k1/k).([NO]²[O2])/[O2]

=k1[NO]²

So, this is not consistent with the rate law.

Mechanism III,

the overall rate =k[NO]².

Therefore, we see that only mechanism I is is most appropriate , reasonable and consistent with the observed rate law.

4 0
3 years ago
An unknown material has a mass of 5.75 g and a volume of 7.5 cm3.
Shtirlitz [24]

Hey there!:

Mass = 5.75 g

Volume = 7.5 cm³

Therefore:

Density = mass / volume

D = 5.75 / 7.5

D = 0.7 g/cm³

Answer C

Hope that helps!

3 0
3 years ago
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