C. the expansion of the universe is accelerating
Taking into account definition of percent yield, the percent yield for the reaction is 76.25%.
<h3>Percent yield</h3>
The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.
The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.
<h3>Percent yield in this case</h3>
In this case, you know:
- actual yield= 1.22 mol
- theorical yield= 1.60 mol
Replacing in the definition of percent yields:

Solving:
<u><em>percent yield= 76.25%</em></u>
Finally, the percent yield for the reaction is 76.25%.
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Answer:
P= 0.87g/mL or 0.87g/cm^3
Explanation:
P=m/v
P=density
P=17.4g/20mL
P= 0.87g/mL
1mL=1cm^3
The mineral galena is composed of lead(II) sulfide and has an average density of 7.46 g/cm³. Moles of lead(II) sulfide in 1.00 ft³ of galena are 883 mol PbS.
<h3>
What is moles of compound?</h3>
A very large number of molecules, atoms, or other particles are referred to as a "mole," which is a unit of measurement in chemistry. The number of moles in one unit is 6.02214 x 1023, and it is known as Avogadro's Number. These figures are crucial in providing information on the quantity of the constituent elements. A mole of a substance might be as little as a few grams or as much as hundreds of grams.
One mole (or formula unit) of a chemical is equal to 6.022 x 10²³ molecules (ionic). A chemical's molar mass represents the mass of 1 mole of that substance. To put it another way, it tells you how many grams there are in a chemical per mole.
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Answer:
Methacrylaldehyde
Explanation:
The first step is the calculation of the <u>IHD</u> (index hydrogen deficiency):


This value indicates that we have <u>2 double bonds</u>. Now, if we check the IR info we can conclude that we have an <u>oxo group</u> (C=O) due to the signal in <u>1705 cm^-1 </u>. So, the options that we can have are <u>aldehyde or ketone</u>.
If we analyze the NMR info we have a signal in 194.7 <u>with only 1 hydrogen</u>. This indicates that necessary we have an <u>aldehyde due to the hydrogen</u>. Also, for the signal in 14 we will have a
, for the signal at 134.2 we will have a
and for the signal at 146.0 we will have a quaternary carbon (no hydrogens present).
So, we will have a
,
, C (without hydrogens), an aldehyde group and a double bond.
When we put all this together we will obtain the <u>Methacrylaldehyde</u> (see figure).