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zheka24 [161]
2 years ago
5

I need help pls! :[​

Chemistry
1 answer:
baherus [9]2 years ago
6 0

dm me on brainly so i can elp you

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Calculate the percent yield when 500 grams of carbon dioxide react with an excess of water to produce 640 grams of carbonic acid
Elena-2011 [213]

Answer:

Percent yield =  90.5%

Explanation:

Given data:

Mass of carbon dioxide = 500 g

Mass of water = excess

Actual yield of carbonic acid = 640 g

Percent yield = ?

Solution:

Balanced chemical equation:

CO₂ + H₂O  → H₂CO₃

Number of moles of carbon dioxide

Number of moles  = Mass / molar mass

Number of moles = 500 g/ 44 g/mol

Number of moles = 11.4 mol

Now we will compare the moles of H₂CO₃ with CO₂.

                              CO₂          :              H₂CO₃

                                 1             :                  1

                               11.4           :                11.4

Mass of carbonic acid:

Mass = number of moles × molar mass

Mass = 11.4 mol × 62.03 g/mol

Mass = 707.14 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield =  640 g/ 707.14 g × 100

Percent yield =  90.5%

7 0
3 years ago
What is a huge organic molecule composed of many smaller molecules called?
photoshop1234 [79]

Answer:

Macromolecules. A very large organic molecule composed of many smaller molecules, 1)Carbohydrates, 2)proteins, 3)lipids, 4)nucleic acids. Three of the four classes of macromolecules that are polymers. 1.Carbohydrates.

7 0
3 years ago
true or false in a chemical reaction the same number of each type of atom will always be present before and after the reaction t
iren [92.7K]

Answer:

True

Explanation:

True because of the law of conservation of mass, the same same amount of atoms will be on both sides of the reaction.

6 0
2 years ago
Consider the elements bromine and chlorine; which element has a larger ionic radius
ipn [44]
Bromine has one more electron shell than the chlorine atom making the radius larger than the chlorine atom.
8 0
3 years ago
What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:
Margarita [4]

Answer:

108.6 g

Explanation:

  • 2NaN₃(s) → 2Na(s) + 3N₂(g)

First we use the <em>PV=nRT formula</em> to <u>calculate the number of nitrogen moles</u>:

  • P = 1.00 atm
  • V = 56.0 L
  • n = ?
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • T = 0 °C ⇒ 0 + 273.2 = 273.2 K

<u>Inputting the data</u>:

  • 1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K
  • n = 2.5 mol

Then we <u>convert 2.5 moles of N₂ into moles of NaN₃</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:

  • 2.5 mol N₂ * \frac{2molNaN_3}{3molN_2} = 1.67 mol NaN₃

Finally we <u>convert 1.67 moles of NaN₃ into grams</u>, using its <em>molar mass</em>:

  • 1.67 mol * 65 g/mol = 108.6 g
6 0
2 years ago
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