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3 years ago

Al+hcl=alcl3+H balance

Chemistry

1 answer:

Mumz [18]3 years ago

8 0

Answer:

Explanation:

Kindly answer the way I have written.

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Ibigay ang kahulugan nito ayun sa paliliwanag<br>konduktor piloto drayber makinista

Westkost [7]

Answer:

konduktor

- ang konduktor ay maaaring isang konduktor nga koryente, train o isang taong namumuno sa isang orkestra o isang grupo ng mga mang-aawit.

piloto

-isang tao na tagalipad ng ereplano

drayber

- isang tao na nagmamaniho ng sasakyan

makinista

-ay isang tao na nagpapakilos o nagpapagana ng makina

please give me brainliest.

6 0

3 years ago

Can someone help me?

kupik [55]

Answer:

#1)2.23hrs

Explanation:

= 134/60

=2.23

=2.23hrs

4 0

3 years ago

How does density of air related to elevation (altitude)

dem82 [27]

Answer:

The density of air goes down as you get higher with altitude.

Explanation:

Less air is pushing down on the molecules below it as you go up. This means that it is not as compacted, another word for which is dense.

8 0

3 years ago

For a particular reaction, Δ=−111.4 kJ and Δ=−25.0 J/K.

vichka [17]

Answer:

$\Delta G =-103.95kJ$

Explanation:

Hello there!

In this case, since the thermodynamic definition of the Gibbs free energy for a change process is:

$\Delta G =\Delta H-T\Delta S$

It is possible to plug in the given H, T and S with consistent units, to obtain the correct G as shown below:

$\Delta G =-111.4kJ-(298K)(-25.0\frac{J}{K}*\frac{1kJ}{1000J} )\\\\\Delta G =-103.95kJ$

Best regards!

6 0

3 years ago

The sulfur dioxide (so2) stack-gas concentration from fossil-fuel combustion is 12 ppmv. determine the stack-gas so2 concentrati

sergejj [24]

The concentration of $SO_{2}$ in the stack gas = 12 ppmv

That means 12 L of $SO_{2}$ is present per $10^{6} L gas$

The given temperature is 273 K (0 C) and pressure is 1 atm. At these conditions, 1 mol of gas would occupy,

$PV = nRT$

$(1 atm) (V) = (1 mol)(0.08206\frac{L.atm}{mol.K}) (273 K)$

V = 22.4 L

1 mol $SO_{2}$ occupies 22.4 L

Moles of $SO_{2}$ = $12 L *\frac{1 mol}{22.4 L} = 0.5357 mol SO_{2}$

Mass of $SO_{2}$ = $0.5357 mol *\frac{64.06 g}{1 mol} = 34.32 g SO_{2} *\frac{10^{6} microgram}{1 g}$ = $3.432 *10^{7}$ μg

Converting $10^{6} L to m^{3}$ :

$10^{6} L *\frac{1 m^{3}}{1000 L}$ = $10^{3} m^{3}$

Calculating the concentration in μg/ $m^{3}$ :

$\frac{3.432 * 10^{7} microgram}{10^{3} L} = 3.432 * 10^{4} microgram/m^{3}$

3 0

3 years ago