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3 years ago

11

Which statements true about the total mass of the reactants during a chemical change? (5 points)

1 answer:

Lapatulllka [165]3 years ago

5 0

It is greater than the total mass

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Please help with question 4 and 5 please!!

ratelena [41]

Question four is : Gaseous state

Question five is : A rise in temperature increases the kinetic energy and speed of particles; it does not weaken the forces between them. The particles in solids vibrate about fixed positions; even at very low temperatures. Individual particles in liquids and gases have no fixed positions and move chaotically.

Hope this helps!

Have a great day!

8 0

3 years ago

Be able to describe the relationship between Freedom of Movement and phase change.

lisabon 2012 [21]

Ion knoe Wdym by “be able to describ’ so ima put it in my own words idr lol:)

if you talm bout some kentic energy or sum ok but other Dan dat ion knoe tbh

I can explain how transferring kinetic energy in and out of a substance can cause a change

7 0

3 years ago

If 0.251moles of H2O gas are produced, how many liters of oxygen gas was used?

Alborosie

Answer:

o.251 prduces 45.7L of oxogen

Explanation:

hope this helps

4 0

3 years ago

Mosquitoes lays eggs that must hatch in water. The young, called larvae live their first few days in water

nevsk [136]

Answer:

Explanation:

I believe he is C

3 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