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Serga [27]
3 years ago
10

How many liters of nitrogen dioxide are in 16 g of NO2 at STP?

Chemistry
1 answer:
mojhsa [17]3 years ago
6 0

Answer: 23 liter

Explanation: Im pretty sure

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Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask. be sure your answer has the correct number
Vera_Pavlovna [14]
<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask. 1mmol = 10^-3 mol Therefore 4.10*10^-5mmol = 4.10*10^-8mol molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below) But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g Mass is = 9.75*10^-7 grams 1µg = 10^-6 g You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4 (*see below) at this point you could have said: 1µg = 10^-6 g therefore you have a solution of 6.29µg per litre, 155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
3 0
3 years ago
10. Isolation of a pure sample of the third product, which has been determined to be an isomer of the major and minor products,
Ymorist [56]

Answer:

Four possible isomers (1–4) for the natural product essramycin. The structure of compound 1 was attributed to essramycin by 1H NMR, 13C NMR, HMBC, HRMS, and IR experiments.

Explanation:

Three synthetic routes were used to prepare all four compounds (Figure 2A). All three reactions utilize 2-(5-amino-4H-1,2,4-triazol-3-yl)-1-phenylethanone (5) as the precursor, whereas each uses different esters (6–8) to construct the pyrimidinone ring. Isomer 1 was prepared by reaction A, which used triazole 5 and ethyl acetoacetate (6) in acetic acid. This was the reaction used in syntheses of essramycin by the Cooper and Moody laboratories.3,4 Reaction B produced compound 2 (minor product) and compound 3 (major product), which were separated chromatographically. This reaction allowed reagent 5 to react with ethyl 3-ethoxy-2-butenoate (7) in the presence of sodium in methanol, under reflux for 24 h. Compound 4 was prepared by reaction C, which was obtained by reflux of 5 and methyl 2-butynoate (8) in n-butanol.

7 0
3 years ago
What is the reaction quotient, Q, for this system when [N2] = 2.00 M, [H2] = 2.00 M, and [NH3] = 1.00 M at 472°C?
cupoosta [38]

Answer : The value of reaction quotient, Q is 0.0625.

Solution : Given,

Concentration of N_2 = 2.00 M

Concentration of H_2 = 2.00 M

Concentration of NH_3 = 1.00 M

Reaction quotient : It is defined as a concentration of a chemical species involved in the chemical reaction.

The balanced equilibrium reaction is,

N_2+3H_2\rightleftharpoons 2NH_3

The expression of reaction quotient for this reaction is,

Q=\frac{[Product]^p}{[Reactant]^r}\\Q=\frac{[NH_3]^2}{[N_2]^1[H_2]^3}

Now put all the given values in this expression, we get

Q=\frac{(1.00)^2}{(2.00)^1(2.00)^3}=0.0625

Therefore, the value of reaction quotient, Q is 0.0625.

3 0
4 years ago
Can a scientific theory turn into a scientific law?
beks73 [17]
Possibly, Yes if it is proven
4 0
4 years ago
Read 2 more answers
What is the vapor pressure at 20 °c of an ideal solution prepared by the addition of 7.38 g of the nonvolatile solute urea, co(n
Romashka [77]

Answer:

83.24 mmHg.

Explanation:

  • <em>The vapor pressure of the solution (Psolution) = (Xmethanol)(P°methanol).</em>

where, Psolution is the vapor pressure of the solution,

Xmethanol is the mole fraction of methanol,

P°methanol is the pure vapor pressure of methanol.

  • We need to calculate the mole fraction of methanol (Xmethanol).

<em>Xmethanol = (n)methanol/(n) total.</em>

where, n methanol is the no. of moles of methanol.

n total is the total no. of moles of methanol and urea.

  • We can calculate the no. of moles of both methanol and urea using the relation: n = mass/molar mass.

n of methanol = mass/molar mass = (56.9 g)/(32.04 g/mol) = 1.776 mol.

n of urea = mass/molar mass = (7.38 g )/(60.06 g/mol) = 0.123 mol.

∴ Xmethanol = (n)methanol/(n) total = (1.776 mol)/(1.776 mol + 0.123 mol) = 0.935.

<em>∴ Psolution = (Xmethanol)(P°methanol)</em> = (0.935)(89.0 mmHg) =<em> 83.24 mmHg.</em>

7 0
3 years ago
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