A chemistry college student is performing a lab exercise wherein they space reacting aqueous command (II) Nitrate with aqueous Potassium Iodide

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

1.57x10^22 atom of Pb

87% yield

Explanation:

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

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

11.4 g PbI2 = 11.4/421 mole = 0.026 moles

expected productivity 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 provided element, the number of neutrons deserve to be different from every other, if the variety of protons is not. These different versions the the same element are dubbed isotopes. Isotopes are atoms with the same number of protons however that have a different number of neutrons.

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Explanation:

inter-base.net:

Scientists to be unsure about the last usual ancestor in between apes and humans due to the fact that fossils room so scarce, researchers perform not recognize what the last usual ancestors of living apes and humans looked favor or wherein they originated.

Explanation:

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

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

Explanation:

To calculate the number of moles, we usage the equation: .....(1)

For 1‑propanol:

Given fixed of 1‑propanol= 8.00 g

Molar mass of 1‑propanol= 60g/mol

Putting worths in equation 1, us get: For oxygen gas:

Given mass of oxygen gas = 49.3 g

Molar fixed of oxygen gas = 32 g/mol

Putting worths in equation 1, us get: The chemistry equation because that the burning of 1-propanol follows: By Stoichiometry the the reaction:

2 moles of 1‑propanol reacts through 9 mole of oxygen gas

So, 0.133 moles of 1‑propanol will certainly react through = of oxygen gas

As, given amount of oxygen gas is more than the forced amount. So, that is taken into consideration as an overabundance reagent.

Thus, 1‑propanol is thought about as a limiting reagent since it boundaries the formation of product.

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

By Stoichiometry the the reaction:

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

So, 0.133 moles of 1‑propanol will create = that carbon dioxide.

Now, calculating the massive of oxygen gas and also carbon dioxide native equation 1, we get:

For oxygen gas:

Molar fixed of oxygen gas = 32 g/mol

Moles that 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 that carbon dioxide = 0.399 moles

Putting worths in equation 1, we get: Hence, the limiting reagent is 1-propanol and the overfill reagent is oxygen gas and also its mass continuing to be is 30.144 grams. The massive of carbon dioxide exit is 17.56 grams