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RUDIKE [14]
3 years ago
14

A soda bottle is flexible enough that the volume of the bottle can change even without opening it. If you have an empty soda bot

tle (volume 591 mL) at room temperature (20 degrees Celsius), what will the new volume be if you put it into a freezer (-4.0 degrees Celsius)?
Chemistry
1 answer:
katen-ka-za [31]3 years ago
3 0

The new volume = 542.6 ml

<h3>Further explanation</h3>

Charles's Law  

<em>When the gas pressure is kept constant, the gas volume is proportional to the temperature </em>

\tt \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

V₁=591 ml

T₁=20+273=293 K

T₂=-4+273=269 K

\tt V_2=\dfrac{V_1.T_2}{T_1}\\\\V_2=\dfrac{591\times 269}{293}\\\\V_2=542.6~ml

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Given w = 0, an endothermic reaction has the following.
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D) + ΔH and +ΔE

Given this is one of the answer choices

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Boron sulfide, B2S3(s), reacts violently with water to form dissolved boric acid (H3BO3) and hydrogen sulfide gas
Gwar [14]

The equation structure for the above mentioned reaction can be written as  

B_{2} S_{3}+6 H_{2} O \rightarrow 2 H_{3} B O_{3}+3 H_{2} S \uparrow

<u>Explanation:</u>

Considering the above reaction, When Boron sulfide, reacts with water more violently to form boric acid and hydrogen sulfide gas.

B_{2} S_{3}+H_{2} O \rightarrow H_{3} B O_{3}+H_{2} S \uparrow

In order to balance the equation, we can do as follows.There are 2 B - atoms on both sides of the equation, but only 2 H - atoms, and one O - atom on LHS, so we have to balance it by putting 6 in front of water and 2 in front of Boric acid and 3 in front of hydrogen sulphide gas, so that we have 2 B - atoms, 3 - S atoms, 12 H - atoms on both sides of the equation, and it is balanced. Balanced equation is given as,

B_{2} S_{3}+6 H_{2} O \rightarrow 2 H_{3} B O_{3}+3 H_{2} S \uparrow

Thus a Balanced equation of the above mentioned reaction is written.

8 0
3 years ago
As the temperature of a liquid increases, the solubility of a liquid in that liquid A. increases. B. decreases. C. stays the sam
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A sample of 0.3283 g of an ionic compound containing the bromide ion (Br−) is dissolved in water and treated with an excess of A
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92.49 %

Explanation:

We first calculate the number of moles n of AgBr in 0.7127 g

n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g

n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol

Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and

From n = m/M

m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol

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% Br in compound = m₁/m₂ × 100%

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m₂ = mass of compound = 0.3283 g

% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %

4 0
3 years ago
What is the volume of 0.200 mol of an ideal gas at 200. kPa and 400. K?
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The ideal gas under STP is 22.4 L/mol. While the gas has a rule of P1V1/T1=P2V2/T2. So the volume under 101 kPa and 273 K is 0.2*22.4=4.48 L.


7 0
3 years ago
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