Explain whether areas B and D show the same season.

_AI + _HCI —> _H2 + _AlCl3

SOVA2 [1]

Hey there!

Al + HCl → H₂ + AlCl₃

Balance Cl.

1 on the left, 3 on the right. Add a coefficient of 3 in front of HCl.

Al + 3HCl → H₂ + AlCl₃

Balance H.

3 on the left, 2 on the right. We have to start by multiplying everything else by 2.

2Al + 3HCl → 2H₂ + 2AlCl₃

Now we have 2 on the right and 4 on the left. Change the coefficient in front of HCl from 3 to 4.

2Al + 4HCl → 2H₂ + 2AlCl₃

Now, for Cl, we have 4 on the left and 6 on the right. Change the coefficient in front of HCl again from 4 to 6.

2Al + 6HCl → 2H₂ + 2AlCl₃

Now, our H is unbalanced again. 6 on the left, 4 on the right. Change the coefficient in front of H₂ from 2 to 3.

2Al + 6HCl → 3H₂ + 2AlCl₃

Balance Al.

2 on the left, 2 on the right. Already balanced.

Here is our final balanced equation:

2Al + 6HCl → 3H₂ + 2AlCl₃

Hope this helps!

7 0

3 years ago

At 570. mm Hg and 25°C, a gas sample ne (in mm Hg) at a volume of 1250 mL and a temperature of 175°C and 25°C, a gas sample has

maw [93]

Answer:

final pressure ( P2) = 467.37 mm Hg

Explanation:

ideal gas:

PV = nRT

∴ P1 = 570 mm Hg * ( atm / 760 mm Hg ) = 0.75 atm

∴ T1 = 25 ° C = 298 K

∴ V1 = 1.250 L

∴ R = 0.082 atm L / K mol

⇒ n = P1*V1 / R*T1

⇒ n = (( 0.75 ) * ( 1.25 )) / (( 0.082 ) * ( 298 ))

⇒ n = 0.038 mol gas

∴ T2 = 175 °C ( 448 K )

∴ V2 = 2.270 L

⇒ P2 = nRT2 / V2

⇒ P2 = (( 0.038 ) * ( 0.082 ) * ( 448 )) / 2.270

⇒ P2 = 0.615 atm * ( 760 mm Hg / atm ) = 467.37 mm Hg

5 0

2 years ago

A gas occupies the volume of 215ml at 15C and 86.4kPa?

kherson [118]

Answer:

About 0.1738 liters

Explanation:

Using the formula PV=nRT, where p represents pressure in atmospheres, v represents volume in liters, n represents the number of moles of ideal gas, R represents the ideal gas constant, and T represents the temperature in kelvin, you can solve this problem. But first, you need to convert to the proper units. 215ml=0.215L, 86.4kPa is about 0.8527 atmospheres, and 15C is 288K. Plugging this into the equation, you get:

$0.8527\cdot 0.215=n \cdot 0.0821 \cdot 288\\n\approx 7.754 \cdot 10^{-3}$

Now that you know the number of moles of gas, you can plug back into the equation with STP conditions:

$1V=7.754 \cdot 10^{-3} \cdot 0.0821 \cdot 273\\V\approx 0.1738L$

Hope this helps!

3 0

3 years ago

PLEASE HELP!! 73 POINTS!!

Reil [10]

Answer:

i am sorry i can not help with that there isnt enough information to work with if you can tell me what you think about the topic i can help you put together a doc

Explanation:

6 0

2 years ago

Which of the following two combinations of reactants is more appropriate for the preparation of p-nitrophenyl phenyl ether? Fluo

VladimirAG [237]

Answer:

p-fluoronitrobenzene and sodium phenoxide is more appropriate

Explanation:

An ipso substitution is required to form p-nitrophenyl phenyl ether.

For this ipso substitution, an alkoxide anion needs to attack as a nucleophile at the carbon atom attached to fluorine atom and thereby substitute that F atom.

p-nitrophenoxide is an weak nucleophile as compared to phenoxide due to presence of electron withdrawing resonating effect of nitro group at para position.

p-fluoronitrobenzene is a good choice for nucleophilic attack by alkoxide anion as compared to fluorobenzene due to higher positive charge density at carbon atom directly attached to F atom. Higher positive charge density arises due to presence of electron withdrawing resonating effect og nitro group at para position.

So, p-fluoronitrobenzene and sodium phenoxide is more appropriate

5 0

3 years ago