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serg [7]
3 years ago
12

If 5.0 liters of carbon dioxide gas are produced by the reaction below at STP, how many liters of oxygen gas were used in the re

action?
2CO (g) + O2 (g) 2CO2 (g)
10.0 L
7.5 L
5.0 L
2.5 L
Chemistry
1 answer:
svp [43]3 years ago
4 0
At STP

1L O₂ → 2L CO₂
xL O₂ → 5.0L CO₂

x=2.5 L
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NemiM [27]

Answer:

The hydrolysis in aqueous HCl of compound A can lead to the formation of a carboxylic acid and an alcohol.

Explanation:

The picture shows the structures of compound A, benzoncaine and the possible products of the proposed reaction.

The acidic hydrolysis is the inverse of the esterification reaction. Therefore, the ester group of compund A will react to form the equivalent carboxylic acid and alcohol.

In order to form benzocaine, the hydrolysis happens in with the nitrile group.

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2 years ago
A bowling ball (mass = 7.2 kg, radius = 0.10 m) and a billiard
evablogger [386]

Answer:

Maximum gravitational Force: F_{gmax} = 1,026*10^{-08} N

Explanation:

The maximum gravitational force is achieved when the center of gravity are the closer they can be. For the spheres the center of gravity is at the center of it, so the closer this two centers of gravity can be is:

bowling ball radius + billiard ball radius = 0,128 m

The general equation for the magnitude of gravitational force is:

F_{gmax}  = G \frac{M*m}{r_{min}^{2} }

Solving for:

G = 6,67*10^{-11}  \frac{Nm^{2}}{kg^{2}}

M = 7,2 kg

m = 0,35 kg

r_{min} = 0,128 m

The result is:

F_{gmax} = 1,026*10^{-08} N

3 0
3 years ago
Carbon-12 and carbon-13 are isotopes of carbon.
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Answer:

Isotopes of an element share the same number of protons but have different numbers of neutrons. Let's use carbon as an example. There are three isotopes of carbon found in nature – carbon-12, carbon-13, and carbon-14. All three have six protons, but their neutron numbers - 6, 7, and 8, respectively - all differ.

Explanation:

4 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.

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