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Naya [18.7K]
3 years ago
6

n Part B, suppose the tablet was mostly dissolved when some of your solution splashed out of the beaker as CO2 continued to evol

ve. How would this affect the perceived mass of CO2? Would your final calculated mass of sodium bicarbonate in the tablet be artificially high or artificially low as a result of this splashing? Choose one, and explain why.
Chemistry
1 answer:
Mandarinka [93]3 years ago
8 0

Answer:

The perceived mass of CO2 would not be affected in large quantities because the splash constitutes small particles of water with sodium bicarbonate that is still reacting. The final calculated mass of sodium bicarbonate in the tablet would be artificially low.

Explanation:

Effervescence is a chemical process that involves the reaction of an acid with a carbonate or sodium bicarbonate, releasing carbon dioxide through a liquid. An example is seen in carbonated beverages, in these the gas that escapes from the liquid is carbon dioxide. The bubbles that are seen are produced by the effervescence of the dissolved gas, which by itself is not visible in its dissolved form.

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hram777 [196]
Protons but different number of neutrons are called isotopes
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4 years ago
Write a transmutation nuclear equation that represents an example of fission
morpeh [17]

Answer:

²³⁸₉₂U →  ²³⁴₉₀Th  + ⁴₂He

Explanation:

Nuclei higher than Bi - 92 naturally are radioactive.

In a transmutation reaction, a new element is produced from an existing one due to radioactivity.

Nuclear fission is the radioactive process by which a heavy nucleus spontaneously decays into lighter ones with the release of a large amount of energy.

 One example is the transmutation of uranium into thorium;

         ²³⁸₉₂U →  ²³⁴₉₀Th  + ⁴₂He

 

4 0
3 years ago
What volume of 12 M HCl solution is needed to make 2.5 L of 1.0 M HCI?
Jlenok [28]

Answer:

0.21 L of 12 M HCL

Explanation:

CV=CV

(12)(x)=(1.0)(2.5)

x=0.21

8 0
4 years ago
How many grams of CO2 will be produced when 8.50 g of methane react with 15.9 g of O2, according to the following reaction? CH4(
Vedmedyk [2.9K]

Taking into account the reaction stoichiometry, 10.93 grams of CO₂ are formed when 8.50 g of methane react with 15.9 g of O₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CH₄ + 2 O₂  → CO₂ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • CH₄: 1 mole
  • O₂: 2 moles
  • CO₂:  1 mole
  • H₂O: 2 moles

The molar mass of the compounds is:

  • CH₄: 16 g/mole
  • O₂: 32 g/mole
  • CO₂:  44 g/mole
  • H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • CH₄: 1 mole ×16 g/mole= 16 grams
  • O₂: 2 moles ×32 g/mole= 64 grams
  • CO₂:  1 mole ×44 g/mole= 44 grams
  • H₂O: 2 moles ×18 g/mole=36 grams

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 16 grams of CH₄ reacts with 64 grams of O₂, 8.50 grams of CH₄ reacts with how much mass of O₂?

mass of O_{2} =\frac{8.50 grams of CH_{4}x64 grams of O_{2} }{16grams of CH_{4}}

mass of O₂= 34 grams

But 34 grams of O₂ are not available, 15.9 grams are available. Since you have less mass than you need to react with 8.50 grams of CH₄, O₂ will be the limiting reagent.

<h3>Mass of CO₂ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 64 grams of O₂ form 44 grams of CO₂, 15.9 grams of O₂ form how much mass of CO₂?

mass of CO_{2} =\frac{15.9 grams of O_{2}x44 grams of CO_{2} }{64grams of O_{2}}

<u><em>mass of CO₂= 10.93 grams</em></u>

Then, 10.93 grams of CO₂ are formed when 8.50 g of methane react with 15.9 g of O₂.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

6 0
2 years ago
How many moles of CO2 are produced when 1 mole wax C31H64 is burned?<br><br> 31 32 64 or 47
Lana71 [14]

Answer:

31 moles

Explanation:

The balanced combustion reaction of the wax, C_{31}H_{64} is shown below as:

C_{31}H_{64}+47O_2\rightarrow 31CO_2+32H_2O

As seen from the reaction,

1 mole of wax, C_{31}H_{64} on combustion produces 31 moles of carbon dioxide, CO_2

<u>Hence, moles of CO_2 when 1 mole of wax, C_{31}H_{64} is burnt = 31 moles</u>

4 0
3 years ago
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