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svet-max [94.6K]
3 years ago
8

Explain why you hear a “whoosh” sound when you open a can containing a carbonated drink. Which gas law applies?

Chemistry
2 answers:
Lana71 [14]3 years ago
6 0

Carbonated drinks have the air under pressure so that carbon bubbles are forced into the drink, keeping it carbonated. So when you open a can, the air under pressure in the can comes out of the can at a high speed, making a "whooshing" sound. The gas law that applies to this concept is the Boyle's Law (PV=k or P1V1=P2V2).

Ad libitum [116K]3 years ago
3 0

Answer:

sorry the other one is a very bad answer i just took the test so this should. good luck

Explanation:

The carbon dioxide in the head space above the liquid is at higher pressure than atmospheric pressure outside the can.

The gas is at a lower volume initially but suddenly has a larger volume available when the can is opened.

The change in pressure as the gas rapidly moves to become dispersed through its new volume causes the “whoosh” sound.

The gas law that applies is Boyle’s law.  

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According to the following reaction, how many grams of potassium phosphate will be formed upon the complete reaction of 29.6 gra
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There is 37.36 grams of K3PO4 produced

Explanation:

Step 1: Data given

Mass of H3PO4 = 29.6 grams

KOH is in excess

Molar mass of KOH = 56.11 g/mol

Molar mass of H3PO4 = 97.99 g/mol

Step 2: The balanced equation

3KOH(aq) + H3PO4(aq) ⇔ K3PO4(aq)+3H2O(l)

Step 3: Calculate mass of KOH

Mass KOH = mass KOH / molar mass KOH

Mass KOH = 29.6 grams / 56.11 g/mol

Mass KOH = 0.528 moles

Step 4: Calculate moles of K3PO4

Since KOH is the limiting reactant, We need 3 moles of KOH for each moles of H3PO4, to produce 1 mole of K3PO4 and 3 moles of H2O

For 0.528 moles of KOH we'll have 0.528/3 =  0.176 moles of K3PO4

Step 5: Calculate mass of K3PO4

Mass K3PO4 = moles K3PO4 * molar mass K3PO4

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PLEASE HELP!!! NEED TO PASS THIS TO THE FIRST SEMESTER!
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Answer:

Reaction 1 is balanced but 2 is not balanced , the balance equation are :

1. CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)

2.CaCl_{2}(aq) + 2NaHCO_{3}(aq)\rightarrow CaCO_{3}(aq) + H_{2}O(l) + 2NaCl(aq) + H_{2}O(aq)

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Balanced Equations : These are the equation which follows the law of conservation of mass .

The total number of atoms present in reactant is equal to total number of atoms present in product.

1. CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)

This is acid - base type reaction where

CH_{3}COOH(aq) act as Acid

NaHCO_{3}(aq) act as weak base

Reactant :CH_{3}COOH(aq) ,NaHCO_{3}

Number of atoms of :

C = 2 (CH_{3}COOH(aq)) + 1 (NaHCO_{3})

   = 2 + 1

   = 3

H  = 4(CH_{3}COOH(aq)) + 1 (NaHCO_{3})

     = 4 + 1

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O = 2(CH_{3}COOH(aq)) + 3 (NaHCO_{3})

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Product :  CO_{2}(g),H_{2}O(l) , CH_{3}COONa(aq)

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C = 1(CO_{2}(g)) + 2(CH_{3}COONa(aq))

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O = 1(H_{2}O(l)) + 2(CH_{3}COONa(aq))

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Na = 1(CH_{3}COONa(aq)

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Check the balancing in both reactant and products should be :

Na = 2

H = 2

Ca = 1

C = 2

O = 6

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