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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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A diver exhales a bubble with a volume of 250 mL at a pressure of 2.4 atm and a temperature of 15 °C. What is the volume of the
alexdok [17]

Answer : The volume of the bubble is, 625 mL

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 2.4 atm

P_2 = final pressure of gas = 1.0 atm

V_1 = initial volume of gas = 250 mL

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 15^oC=273+15=288K

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Now put all the given values in the above equation, we get:

\frac{2.4atm\times 250mL}{288K}=\frac{1.0atm\times V_2}{300K}

V_2=625mL

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7 0
2 years ago
An atom of gold has a mass of 3.271 X 10-22 g. How many atoms of gold are in 5.00 g of gold? (Give your answer in scientific not
Kobotan [32]

Answer:

1.53 × 10²² atoms Ag

Explanation:

Step 1: Define conversions

3.271 × 10⁻²² g = 1 atom

Step 2: Use Dimensional Analysis

5.00 \hspace{3} g \hspace{3} Ag(\frac{1 \hspace{3} atom \hspace{3} Ag}{3.271(10)^{-22} \hspace{3} g \hspace{3} Ag} ) = 1.52858 × 10²² atoms Ag

Step 3: Simplify

We have 3 sig figs.

1.52858 × 10²² atoms Ag ≈ 1.53 × 10²² atoms Ag

5 0
2 years ago
What is the freezing point depression of a solution that contains 0.705?
Sveta_85 [38]
Colligative properties calculations are used for this type of problem. Calculations are as follows:<span>
</span>

<span>ΔT(freezing point)  = (Kf)m
ΔT(freezing point)  = 1.86 °C kg / mol (0.705)
ΔT(freezing point)  = 1.3113 °C

</span>

<span>
</span>

<span>Hope this answers the question. Have a nice day.</span>

6 0
3 years ago
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