A chemistry student is performing a lab exercise where they are reacting aqueous Lead (II) Nitrate with aqueous Potassium Iodide

inter-base.net: Pb(NO3)2(aq) + 2KI(aq) —> PbI2(s) + 2KNO3(aq)

1.57x10^22 atoms of Pb

87% yield

Explanation:

10.0 g Pb(II) nitrate = 10/333 moles = 0.03 moles

12.0 g KI = 12/166 moles = 0.072 moles (0.06 moles required)

11.4 g PbI2 = 11.4/421 moles = 0.026 moles

expected yield 0.03 moles, yield = 87%

atoms in 11.4 g = 0.026 moles = 0.026*6.02214076*10^23 atoms = 1.57*10^22



inter-base.net:

In a given element, the number of neutrons can be different from each other, while the number of protons is not. These different versions of the same element are called isotopes. Isotopes are atoms with the same number of protons but that have a different number of neutrons.

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Explanation:



inter-base.net:

Scientists were unsure about the last common ancestor between apes and humans because fossils are so scarce, researchers do not know what the last common ancestors of living apes and humans looked like or where they originated.

Explanation:




A sample of 8.00 g of liquid 1‑propanol, C 3 H 8 O , is combusted with 49.3 g of oxygen gas. Carbon dioxide and water are the pr

inter-base.net: The limiting reagent is 1-propanol and the excess reagent is oxygen gas and its mass remaining is 30.144 grams. The mass of carbon dioxide released is 17.56 grams

Explanation:

To calculate the number of moles, we use the equation:

*
.....(1)

For 1‑propanol:

Given mass of 1‑propanol= 8.00 g

Molar mass of 1‑propanol= 60g/mol

Putting values in equation 1, we get:

*

For oxygen gas:

Given mass of oxygen gas = 49.3 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

*

The chemical equation for the combustion of 1-propanol follows:

*

By Stoichiometry of the reaction:

2 moles of 1‑propanol reacts with 9 moles of oxygen gas

So, 0.133 moles of 1‑propanol will react with =

*
of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, 1‑propanol is considered as a limiting reagent because it limits the formation of product.

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Moles of excess reagent (oxygen gas) = <1.54 - 0.598> = 0.942 moles

By Stoichiometry of the reaction:

2 moles of 1‑propanol produces 6 moles of carbon dioxide

So, 0.133 moles of 1‑propanol will produce =

*
of carbon dioxide.

Now, calculating the mass of oxygen gas and carbon dioxide from equation 1, we get:

For oxygen gas:

Molar mass of oxygen gas = 32 g/mol

Moles of oxygen gas = 0.942 moles

Putting values in equation 1, we get:

*

For carbon dioxide gas:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.399 moles

Putting values in equation 1, we get:

*

Hence, the limiting reagent is 1-propanol and the excess reagent is oxygen gas and its mass remaining is 30.144 grams. The mass of carbon dioxide released is 17.56 grams