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77julia77 [94]
3 years ago
6

Hdhdhzjzjzj which??????????????????

Chemistry
1 answer:
dsp733 years ago
8 0

Answer:

I think it’s D but not sure

Explanation:

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What is the reaction scheme for the experiment of synthesis of methyl salicylate
soldier1979 [14.2K]

The methyl salicylate (wintergreen oil) is the methyl ester of salicylic acid.

Methyl salicylate is produced by esterifying salicylic acid with methanol, the reaction scheme is attached to this answer. In the past, it was commonly distilled from the twigs of Betula lenta (sweet birch) and Gaultheria procumbens (eastern teaberry or wintergreen).

3 0
3 years ago
In the laboratory you dissolve 12.3 g of chromium(ii) sulfate in a volumetric flask and add water to a total volume of 375. ml.
valentinak56 [21]
Hope it cleared your doubt.

4 0
3 years ago
How many liters of nitrogen gas are needed to make 25 mol of nitrogen trifluoride
ikadub [295]
The reaction between N₂ and F₂ gives Nitrogen trifluoride as the product. The balanced equation is;

N₂ + 3F₂ → 2NF₃

The stoichiometric ratio between N₂ and NF₃ is 1 : 2
Hence, 
  moles of N₂ / moles of F₂ = 1 / 2
  moles of N₂ / 25 mol        =  0.5
           moles of N₂            =  0.5 x 25 mol = 12.5 mol
 
Hence N₂ moles needed  = 12.5 mol

At STP (273 K and 1 atm) 1 mol of gas = 22.4 L

Hence needed N₂ volume = 22.4 L mol⁻¹ x 12.5 mol
                                          = 280 L 
6 0
3 years ago
Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow
andreyandreev [35.5K]

Explanation:

(a)  The given data is as follows.

              Pressure on top (P_{o}) = 140 bar = 1.4 \times 10^{7} Pa       (as 1 bar = 10^{5})

              Temperature = 15^{o}C = (15 + 273) K = 288 K

         Density of gas = \frac{PM}{ZRT}

                \frac{dP}{dZ} = \rho \times g

               \frac{dP}{dZ} = \int \frac{PM}{ZRT}

                \int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ

           ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}

                              = 0.4548

                     P_{1} = P_{o} \times e^{0.4548}

                                 = 1.4 \times 10^{7} Pa \times 1.5797

                                 = 2.206 \times 10^{7} Pa

Hence, pressure at the natural gas-oil interface is 2.206 \times 10^{7} Pa.

(b)   At the bottom of the tank,

                 P_{2} = P_{1}  + \rho \times g \times h

                             = 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

                             = 309.8 \times 10^{5} Pa

                             = 309.8 bar

Hence, at the bottom of the well at 15^{o}C pressure is 309.8 bar.

6 0
3 years ago
What is the mass of a rectangular cube with dimensions of 3 cm x 2 cm x to 10 cm and a density of 10 g/cm^3
marta [7]

d=m/V so m=Vd

m=60cm^3 * 10g/cm^3=600g

3 0
3 years ago
Read 2 more answers
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