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2 years ago

12

What is Lewis acid and Lewis base? give examples

1 answer:

AleksAgata [21]2 years ago

6 0

Explanation:

example is copper iron...........

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Which of the following you do NOT need to pay attention while using a centrifuge?

Phoenix [80]

Answer:

The radius of the centrifuge.

Explanation:

Hello,

Since the radius of the centrifuge is just a design parameter, it wouldn't be a cause of failure because it is used to know how many tubes could be fitted in into the centrifuge. On the other hand, keeping you attention away from other factors could turn into a failure as long as the sample could be poured down or just turn out inadequate for the expected results.

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3 years ago

The Dust Bowl was a time In the 1930s when prairie states in the Midwest had farmlands torn apart by wind erosion and storms. Th

grandymaker [24]

D. I might be wrong but I learned this a while ago

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2 years ago

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I need help on science lol.

Anit [1.1K]

Answer:

yes your right its is 4

Explanation:

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2 years ago

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A student carefully pipets 10.00ml of ethanol into a beaker. If ethanol has a density of 0.779g/ml, how many grams of ethanol we

lesantik [10]

Answer: 7.79 grams of ethanol were put into the beaker.

Explanation:

To calculate the mass of ethanol, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of ethanol = 0.779 g/mL

Volume of water = 10.00 mL

Putting values in above equation, we get:

$0.779g/mL=\frac{\text{Mass of ethanol}}{10.00mL}\\\\\text{Mass of ethanol}=(0.779g/mL\times 10.00mL)=7.79g$

Thus 7.79 grams of ethanol were put into the beaker.

5 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