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NeTakaya
3 years ago
11

given the mass lost during the reaction when the alka seltzer is the limiting reagent and that the molar mass of carbon dioxide

is 44.01 g/mol, how many moles of CO2 were produced
Chemistry
1 answer:
Vadim26 [7]3 years ago
5 0
Alka-seltzer is an effervescent antacid which neutralizes the acid of the stomach by generating CO2. The chemical formula of alka-seltzer is C16H17NaO14. In this case, if we are given the mass of the antacid used, then by stoichiometry, mass alka-seltzer * (1/ molar mass of alk-selz) * (1 mol Co2 / 1 mol alk-seltz) * (44 g/molCo2). That is how you get the mass of Co2 produced.
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Answer:

<em>In equipment reliability, the likelihood that a given event will occur</em>

Explanation:

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Decide which element probably has a boiling point most and least similar to the boiling point of cesium.
natka813 [3]

Answer:

Take a look at the attachment below

Explanation:

Take a look at the periodic table. As you can see, Rubidium is the closest element to Cesium, and happens to have the closest boiling point to Cesium, with only a difference of about 30 degrees.

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<em>Hope that helps!</em>

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Please help with 14-16 <br> Thank you!
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4 0
2 years ago
An example of a physical property of an element is the element's ability to
FromTheMoon [43]

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4 0
2 years ago
Read 2 more answers
Calculate the freezing point and boiling point of a solution containing 8.15 g of ethylene glycol (C2H6O2) in 96.3 mL of ethanol
pishuonlain [190]

<u>Answer:</u> The freezing point of solution is -117.54°C and the boiling point of solution is 80.48°C

<u>Explanation:</u>

To calculate the mass of ethanol, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of ethanol = 0.789 g/mL

Volume of ethanol = 96.3 mL

Putting values in above equation, we get:

0.789g/mL=\frac{\text{Mass of ethanol}}{96.3mL}\\\\\text{Mass of ethanol}=(0.789g/mL\times 96.3mL)=75.98g

  • <u>Calculating the freezing point:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = -114.1 °C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 1.99°C/m

m_{solute} = Given mass of solute (ethylene glycol) = 8.15 g

M_{solute} = Molar mass of solute (ethylene glycol) = 62 g/mol

W_{solvent} = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

-114.1-\text{Freezing point of solution}=1\times 1.99^oC/m\times \frac{8.15\times 1000}{62g/mol\times 75.98}\\\\\text{Freezing point of solution}=-117.54^oC

Hence, the freezing point of solution is -117.54°C

  • <u>Calculating the boiling point:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

Boiling point of pure solution = 78.4°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_b = molal boiling point elevation constant = 1.20°C/m.g

m_{solute} = Given mass of solute (ethylene glycol) = 8.15 g

M_{solute} = Molar mass of solute (ethylene glycol) = 62  g/mol

W_{solvent} = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

\text{Boiling point of solution}-78.4=1\times 1.20^oC/m\times \frac{8.15\times 1000}{62\times 75.98}\\\\\text{Boiling point of solution}=80.48^oC

Hence, the boiling point of solution is 80.48°C

3 0
3 years ago
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