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3 years ago

14

For the reaction H2SO4 + 2Na2OH4 —> Na2SO4 + 2H2O

1 answer:

beks73 [17]3 years ago

5 0

The answer is b) 12 i believe

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Isotopes are atoms of an element

stellarik [79]

D. With the same number of protons and different number of neutrons.

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3 years ago

Calculate the energy of a photon with a frequency of 9.78 x 1021 Hz.

Alex_Xolod [135]

Answer:

9.98538 kilohertz

Explanation:

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8 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

Why was mass lost from the crucible during the reaction?

Alexxandr [17]

If the crucible wasn't covered with a lid the reactants may have produced a gas that was released into the surroundings, or mass may have been lost in the form of water vapour.

6 0

3 years ago

A gas balloon has a volume of 106.0 liters when the temperature is 25.0 °C and the pressure is 740.0 mm Hg. What will its volume

ipn [44]

Answer : The final volume of gas will be, 103.3 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 740.0 mmHg = 98.4 kPa

Conversion used : (1 mmHg = 0.133 kPa)

$P_2$ = final pressure of gas = 99.3 kPa

$V_1$ = initial volume of gas = 106.0 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $25.0^oC=273+25.0=298K$

$T_2$ = final temperature of gas = $20.0^oC=273+20.0=293K$

Now put all the given values in the above equation, we get:

$\frac{98.4kPa\times 106.0L}{298K}=\frac{99.3kPa\times V_2}{293K}$

$V_2=103.3L$

Therefore, the final volume of gas will be, 103.3 L

8 0

3 years ago