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3 years ago

Net ionic equation for Barium chloride

Chemistry

1 answer:

vaieri [72.5K]3 years ago

3 0

Answer: There is one way to write it but i’ll also provide an unbalanced equation and a balanced one.

Explanation:

Unbalanced : Ba (aq) + Cl2 (aq)—-> BaCl (aq)

Balanced : 2Ba (aq) + Cl2 (aq)—> 2BaCl(aq)

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2 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

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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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4 years ago

What volume of 0.128 M HCl is needed to neutralize 2.87 G of Mg(OH)2?

qaws [65]

V= .768L= 76.8ml HCL

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4 years ago

Net ionic equation for Barium chloride​

Net ionic equation for Barium chloride