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3 years ago

9

I NEED HELP PLEASE, THANKS! :)

1 answer:

konstantin123 [22]3 years ago

7 0

Answer:

Gasoline isn't a pure substance.

Explanation:

A typical boiling range for gasoline is 100–400F. Gasoline has a vapor pressure of 7 psi in the summer, comparing it to water pressure it's less then 0.3 psi. For example if you spill water on a flat surface it might take a few minutes or hours to completly evapourate. Gasoline on the other hand it will evapurate within a few minutes, that's an example of volatility. Hope this helps

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Which substance is a primary source of hydrocarbon

IceJOKER [234]

Answer:

Hydrocarbons are molecules consisting of both hydrogen and carbon. They are most famous for being the primary constituent of fossil fuels, namely natural gas, petroleum, and coal. For this reason, fossil fuel resources are often referred to as hydrocarbon resources.Jun 25, 2018

https://energyeducation.ca ›

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3 years ago

OK! this is for the weebs :D

maw [93]

Answer:

Explanation:3

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3 years ago

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What are the products in the chemical equation for the overall process of photosynthesis?

Alik [6]

Glucose and a plants and ur welcome

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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

Consider the formation of hcn by the reaction of nacn (sodium cyanide) with an acid such as h2so4 (sulfuric acid): 2nacn(s)+h2so

Artemon [7]

2NaCN(s) + H₂SO₄(aq) --> Na₂SO₄(aq) + 2HCN(g)

The molar ratio between NaCN : HCN is 2:2 or 1:1

Mass of HCN = 16.7 g

Molar mass of HCN = 1 + 12 + 14 = 27 g/mol

Molar mass of NaCN = 49 g/mol

Therefore, the mass of NaCN is

16.7 g of HCN x 49 g/mol of NaCN / 27 g/mol of HCN = 30.3 grams of NaCN

Therefore, 30.3 grams of NaCN gives the lethal dose in the room.

5 0

3 years ago

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