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il63 [147K]
1 year ago
10

What is the composition of stainless steel​

Chemistry
2 answers:
timofeeve [1]1 year ago
4 0

Stainless steel is composed of:

73% Iron

18% Chromium

8% Nickel

1% Carbon

andreev551 [17]1 year ago
3 0
Stainless steel is an alloy of iron, chromium and nickel.
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The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=k[H2][I2] Some proposed mechanism
MatroZZZ [7]

Answer:

Mechanism A and B are consistent with observed rate law

Mechanism A is consistent with the observation of J. H. Sullivan

Explanation:

In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.

In the proposed mechanisms:

Mechanism A

(1) H2(g)+I2(g)→2HI(g)(one-step reaction)

Mechanism B

(1) I2(g)⇄2I(g)(fast, equilibrium)

(2) H2(g)+2I(g)→2HI(g) (slow)

Mechanism C

(1) I2(g) ⇄ 2I(g)(fast, equilibrium)

(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)

(3) H(g)+I(g)→HI(g) (fast)

The rate laws are:

A: rate = k₁ [H2] [I2]

B: rate = k₂ [H2] [I]²

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]

<em>Where K' = K1 * K2</em>

C: rate = k₁ [H2] [I]

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]^1/2

Thus, just <em>mechanism A and B are consistent with observed rate law</em>

In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just <em>mechanism A is consistent with the observation of J. H. Sullivan</em>

5 0
3 years ago
What inference can be drawn from the graph?
defon

Answer:

i think its C

Explanation:

7 0
1 year ago
If the reactants are 10 kg what is the mass of the products
nlexa [21]

If the mass of both the reactants is 10kg then the mass of the products also equals 10kg.

It is due to the law of conservation of mass.

Mass can neither be created nor be destroyed.

7 0
2 years ago
Which of the following is the last step in performing a titration
Alinara [238K]
The final step in a typical titration, that is here an acid base one would be to finally find the concentration of your unknown substance whether that be the acid or the base. The other steps are used before this to come to the correct calculation and conclusion.
7 0
2 years ago
Read 2 more answers
Using Equation (10), calculate [Ag+] in the cell, where it is in equilibrium with 1 M Cl- ion. (Ecell in Equation (10) is the ne
mixer [17]

Answer:

7.16x10⁻⁸M = [Ag+]

Explanation:

Using the equation:

E(Cell) =E⁰ - 0.0592/2 • log ([Cu2+]/[Ag+]²)

<em>Where E</em>⁰<em>= 0.4249V</em>

<em>E(Cell) = -(-0.0019V) -Measured value-</em>

<em>[Cu2+] = 1M</em>

<em />

Replacing:

0.0019V = 0.4249V - 0.0592/2 • log (1M/[Ag+]²)

-0.423V = - 0.0296 • log (1M/[Ag+]²)

14.29 = log (1M/[Ag+]²)

1.95x10¹⁴ = 1M / [Ag+]²

[Ag+]² = 5.12x10⁻¹⁵M

7.16x10⁻⁸M = [Ag+]

5 0
2 years ago
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