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3 years ago

If 97.3 L of NO2 forms measured at 35 C and 632 mm Hg. What is the percent yield?

1 answer:

3 0

a. Use PV = nRT and solve for n = number of mols O2.
mols NO = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one is wrong; the correct value in LR (limiting reagent) problems is ALWAYS the smaller value and the reagent producing that value is the LR.

b.
Using the smaller value for mols NO2 from part a, substitute for n in PV = nRT, use the conditions listed in part b, and solve for V in liters. This will give you the theoretical yield (YY)in liters. The actual yield at these same conditions (AY) is 84.8 L.

and % will be 60%.

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Which of the following defines the wavelength of a wave?

Makovka662 [10]

the distance between peaks of a wave

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4 years ago

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CaCO3 + 2HCl --> CaCl2 + CO2 + H2O In the reaction above, what mass of carbon dioxide is produced when 10 moles of HCl reacts

puteri [66]

According to the stoichiometry the CO2 formed is 1/2 the moles of H2O. So take the moles of CO2 as 5moles.

n=m/M

Where :

n = mole

m = mass

M = molar mass

Molar mass of CO2 = 44gmol^-1

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4 years ago

A block of material has a volume of 50 cubic centimeters, and a mass of 1,000 grams; what is that object's density?

zhannawk [14.2K]

Answer:

Density = $20\ g/cm^3$

Explanation:

Given that,

Mass of the object, m = 1000 g

Volume of the block, V = 50 cm³

We need to find the density of the object. Density is equal to mass per unit volume.

d = m/V

$d=\dfrac{1000\ g}{50\ cm^3}\\\\=20\ g/cm^3$

So, the density of the object is $20\ g/cm^3$ .

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3 years ago

5. How many grams of potassium bromide will dissolve in 450g of water at 80°C?

Shtirlitz [24]

Hi mate. I think the answer is 100

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3 years ago

Which of the following substances would you predict to have the highest DHvap? Group of answer choices CH3CH2CH2CH3 CH3CH2OH HF

Katyanochek1 [597]

Answer:

a) $CH_3CH_2CH_2CH_3$

Explanation:

In this question we have the following answer choices:

a) $CH_3CH_2CH_2CH_3$

b) $CH_3CH_2OH$

c) $HF$

d) $CH_3Cl$

e) $HOCH_2CH_2OH$

We have to remember the relationship between intermolecular forces and vapor pressure. If we have stronger intermolecular forces we will have less vapor pressure because the molecules have more interactions between them, so, the molecules will prefer to stay in a liquid state rather than a gaseous state. Now, we have to check each molecule:

a) $CH_3CH_2CH_2CH_3$ (Van der waals interactions)

b) $CH_3CH_2OH$ (Hydrogen bonding)

c) $HF$ (Hydrogen bonding)

d) $CH_3Cl$ (Dipole-dipole interaction)

e) $HOCH_2CH_2OH$ (Hydrogen bonding)

For molecules b, c and e we have hydrogen bond to a heteroatom (O, N, S, or P). In this case oxygen, therefore we will have hydrogen bonding interactions (a very strong interaction). So, we can discard these ones.

In molecule e, we have "Cl" bond to a "C" therefore we will have the presence of a dipole (due to the electronegativity difference). If we have a dipole, we will have a dipole-dipole interaction (a strong interaction, less than hydrogen bonding but still is a strong interaction).

In molecule a, we have only Van der Waals interactions because in this molecule we have only carbon and hydrogen atoms bonded by single bonds. So, we will have a non-polar molecule. These interactions are the weakest interactions of all the molecules given. So, if we have weaker interactions the molecules can be converted to a gas state more easily and we have more vapor pressure. 

7 0

3 years ago