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svp [43]
3 years ago
9

Could use some help on this one

Chemistry
1 answer:
cluponka [151]3 years ago
3 0
Sorry for the sketchy drawing.

b. this is an alpha 1-6 glycosidic linkage

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Is water wet? My friends are debating.​
romanna [79]

Answer:

No

Explanation:

<em>Water itself is not wet, but when its touches something dry it makes the dry object wet.</em>

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3 years ago
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Which of the following is an example of an energy transformation?
erica [24]

Answer:

D

Explanation:

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3 years ago
How many atoms of hydrogen are in 0.500 mol of ch3oh molecules?
cestrela7 [59]
In 1 mol of CH3OH, you have 4 H-atoms (because 3 H-atoms are attached to the C-atom, and one H-atom in the OH group). That means in 0.500 mol of CH3OH, you have 2 H-atoms since it is halved. And then we have Avogadro's constant: 6.02 * 1023.

The question asks for how many hydrogen atoms there are in 0.500 mol CH3OH. Using the numbers that we have (Avogadro's constant and no. of H-atoms), the answer of the question will be something like:

<span>H-atoms in CH3OH = 2 * 6.02 * </span>1023<span> = ~1.2 * 10</span>24

 


8 0
3 years ago
Is na2o a element or compound
Orlov [11]

Answer:

Its a compound :PP

Explanation:

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5 0
3 years ago
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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
4 years ago
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