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Vesna [10]
3 years ago
10

A typical sheet of weighing paper is 3.5 inches by 4.1 inches and weighs 0.45 g. What is its absorber thickness in g/cm^{2} 2 

?
Chemistry
1 answer:
SpyIntel [72]3 years ago
6 0

Answer:

\boxed{\textbf{0.0046 g/cm}^{2}}

Explanation:

1. Convert inches to centimetres

l= \text{4.1 in} \times \dfrac{\text{2.54 cm}}{\text{1 in}} = \text{10.2 cm}\\\\w= \text{3.5 in} \times \dfrac{\text{2.54 cm}}{\text{1 in}} = \text{8.89 cm}

2. Calculate the area

A = lw = 10.2 × 8.89= 90.3 cm²

3. Calculate the absorber thickness

\text{Absorber thickness} = \dfrac{\text{0.415 g}}{\text{90.3 cm}^{2}} = \textbf{0.0046 g/cm}^{2}\\\\\text{The absorber thickness is $\boxed{\textbf{0.0046 g/cm}^{2}}$}

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<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)

Rate law expression for the reaction:

\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b

where,

a = order with respect to HgCl_2

b = order with respect to C_2O_4^{2-}

Expression for rate law for first observation:

3.2\times 10^{-5}=k(0.164)^a(0.15)^b  ....(1)

Expression for rate law for second observation:

2.9\times 10^{-4}=k(0.164)^a(0.45)^b  ....(2)

Expression for rate law for third observation:

1.4\times 10^{-4}=k(0.082)^a(0.45)^b  ....(3)

Expression for rate law for fourth observation:

4.8\times 10^{-5}=k(0.246)^a(0.15)^b  ....(4)  

Dividing 2 from 1, we get:

\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2

Dividing 2 from 3, we get:

\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2

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