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kondor19780726 [428]
3 years ago
13

What is the formula for aluminum sulfate?

Chemistry
1 answer:
Bas_tet [7]3 years ago
3 0
Al2(SO4)<span>3 is the compound formula

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If 725 ml of N² gas is at 1 atm, what volume will it have at 1.4 atm?​
alexdok [17]

Answer:

518 mL

Explanation:

We can solve this using Boyle's Law Formula

P1V1 = P2V2

where p1 = initial pressure, p2 = final pressure, v1 = initial volume and v2 = final volume

here , the initial pressure is 1 atm and the initial volume is 725mL

we are given the final pressure 1.4 and we need to find the final volume

so we have p1v1 = p2v2

==> plug in p1 = 1 , v1 = 725 mL and p2 = 1.4

(1)(725) = (1.4)v2

==> multiply 1 and 725

725 = (1.4)(v2)

==> divide both sides by 1.4

v2 = 518

N2 would have a volume of 518mL at 1.4atm

7 0
2 years ago
Read 2 more answers
What is 5L to mL ( chemistry )
zlopas [31]

There are 1000 mililiters in a liter, so 1000 ml for every liter, you have 5 liters, so:

5L*1000 = 5000 mL

5 0
3 years ago
Read 2 more answers
How many grams of glucose are in 11.5 moles?
mestny [16]

There are 2071.79 grams of glucose in 11.5 moles.

8 0
3 years ago
Calculate the molarity of 48.0 mL of 6.00 M H2SO4 diluted to 0.250 L
const2013 [10]

Answer:

The answer is 1.15m.

Since molality is defined as moles of solute divided by kg of solvent, we need to calculated the moles of H2SO4 and the mass of the solvent, which I presume is water.

We can find the number of H2SO4 moles by using its molarity

C=nV→nH2SO4=C⋅VH2SO4=6.00molesL⋅48.0⋅10−3L=0.288

Since water has a density of 1.00kgL, the mass of solvent is

m=ρ⋅Vwater=1.00kgL⋅0.250L=0.250 kg

Therefore, molality is

m=nmass.solvent=0.288moles0.250kg=1.15m

4 0
3 years ago
Read 2 more answers
Purification of nickel can be achieved by electrorefining nickel from an impure nickel anode onto a pure nickel cathode in an el
Alexxandr [17]

Answer: 530 hours

Explanation:

The reduction of Nickel ions to nickel is shown as:

Ni^{2+}+2e^-\rightarrow Ni

96500\times 2=193000Coloumb of electricity deposits 1 mole of Nickel

1 mole of Nickel weighs = 58.7 g

Given quantity = 18.0 kg = 18000 g  (1kg=1000g)

58.7 g of Nickel is deposited by 193000 C of electricity

18000 g of Nickel is deposited by = \frac{193000}{58.7}\times 18000=59182282.8C of electricity

Q=I\times t

where Q= quantity of electricity in coloumbs  = 59182282.8C

I = current in amperes = 31.0 A

t= time in seconds = ?

59182282.8C=31.0A\times t

t=1909105.9sec

(1h=3600 sec)

t=530hours

Thus 530 hours are required to plate 18.0 kg of nickel onto the cathode if the current passed through the cell is held constant at 31.0 A

3 0
3 years ago
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