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8_murik_8 [283]
3 years ago
6

Nuclear equations 1/0n + / = 236/92 U

Chemistry
1 answer:
andre [41]3 years ago
7 0

Answer:

I think this may help

Explanation:

Sum of subscripts on left = 92. Sum of subscripts on right = 57. So X must have atomic number = 92 – 57 = 35. Element 35 is bromine.

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Based on the information that is given, which atom is the table ?
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Read 2 more answers
Can someone help me with 2 and 3? I don't know the answer and I need justifications for why the correct answers are correct
saw5 [17]

Answer:

2. (C) K < 1.

3. (B) [Fe³⁺] = 2.00 mol·L⁻¹; [SCN⁻] = 6.0 mol·L⁻¹

Step-by-step explanation:

2. Value of K

A⇌ B

K = [B]/[A]

If the concentration of reactants (A) is larger than the concentration of products (B), the denominator of the K expression is larger than the numerator.

The fraction is less than 1, so

K < 1

3. Equilibrium concentrations

We can use an ICE table to keep track of the calculations.

                     Fe³⁺ + SCN⁻ ⇌ FeSCN²⁺

I/mol·L⁻¹:     6.00      10.0            0

C/mol·L⁻¹:     -x           -x             +x

E/mol·L⁻¹:  6.00-x  10.0-x            x

Initially, there is no FeSCN²⁺ present, so [FeSCN²⁺] = 0.

Then, the Fe³⁺ and SCN⁻ react until equilibrium is reached.

How much will react? We don't know, but we have every confidence that x mol (some unknown quantity) will react.

[Fe³⁺] will <em>decrease</em> by x mol·L⁻¹. Because of the 1:1:1 molar ratios, [SCN⁻] will also <em>decrease</em> by x mol·L⁻¹ and [FeSCN²⁺] will <em>increase</em> by x mol·L⁻¹.

We add the changes and get the values in the bottom line.

However, what is the value of that pesky x?

We are told that [FeSCN²⁺] = 4.00 mol·L⁻¹ at equilibrium.

From the table, x = [FeSCN²⁺], so x = 4.00.

Now we can insert these values back into the table.

At equilibrium,

[Fe³⁺]   =   6.00 - x =  6.00 - 4.00 = 2.00 mol·L⁻¹

[SCN⁻] = 10.0    - x = 10.0   - 4.00 = 6.0   mol·L⁻¹

5 0
2 years ago
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