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ipn [44]
2 years ago
5

An experiment with 55 co takes 47.5 hours. at the end of the experiment, 1.90 ng of 55-co remains. if the half-life is 18.0 hour

s, how many ng of 55-co were originally present?
Chemistry
1 answer:
Andru [333]2 years ago
7 0

Answer:

\boxed{\text{10.7 ng}}

Explanation:

Let A₀ = the original amount of ⁵⁵Co .

The amount remaining after one half-life is ½A₀.

After two half-lives, the amount remaining is ½ ×½A₀ = (½)²A₀.

After three half-lives, the amount remaining is ½ ×(½)²A₀ = (½)³A₀.

The general formula for the amount remaining is:

A =A₀(½)ⁿ

where n is the number of half-lives

n = t/t_½

Data:

   A = 1.90 ng

    t = 45 h

t_½ = 18.0 h

Calculation:

(a) Calculate n

n = 45/18.0 = 2.5

(b) Calculate A

1.90 = A₀ × (½)^2.5

1.90 = A₀ × 0.178

A₀ = 1.90/0.178 = 10.7 ng

The original mass of ⁵⁵Co was \boxed{\text{10.7 ng}}.

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Answer:

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Explanation:

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Use Lewis theory to determine the chemical formula for the compound formed between Mg and Br. Group of answer choices Mg3Br2 Mg2
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(Mg^2+) + (Br^1-)

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3 years ago
You wish to make 250. mL of 0.20 M KCl from a stock solution of 1.5 M KCl and DI water.
Nitella [24]

The question requires us to use the dilution formula M_iV_i=M_fV_f, where M_i and V_i are the stock concentration and volume respectively, then M_f and V_f are the dilute concentration and volume respectively.

a. C_s_t_o_c_k= 1.5 M KCI, C_d_i_l_u_t_e=0.20M KCl, V_d_i_l_u_t_e=250ml KCl

b.M_iV_i=M_fV_f\\\implies V_i= \frac{M_fV_f}{M_i} = \frac{0.20M \times 250ml}{1.5M} = 33.3\ ml

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8 0
3 years ago
What is the empirical formula for a substance that is composed of 40.66% carbon, 8.53% Hydrogen,23.72% Nitrogen, and 27.09% Oxyg
prohojiy [21]

Answer:

THE EMPIRICAL FORMULA OF THE SUBSTANCE IS C2H5NO

Explanation:

The steps involved in calculating the empirical formula of this substance in shown in the table below:

Element                            Carbon           Hydrogen          Nitrogen         Oxygen

1. % Composition              40.66              8.53                 23.72                27.09

2. Mole ratio =

%mass/ atomic mass       40.66/12         8.53/1          23.72/14            27.09/16

                                       =  3.3883            8.53              1,6943             1.6931

3. Divide by smallest

value (0.6931)          3.3883/1.6931    8.53/1.6931    1.6943/1.6931   1.6931/1.6931

                               =      2.001                  5.038           1.0007                      1

4. Whole number ratio        2                       5                   1                               1

The empirical formula = C2H5NO

7 0
3 years ago
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