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astraxan [27]
3 years ago
7

What is the gas produced when iron(II) sulfide is treated with a nonoxidizing acid?

Chemistry
1 answer:
Art [367]3 years ago
3 0

Answer:

Hydrogen sulfide

Explanation:

A nonoxidizing acid is the acid which cannot act as oxidizing agent and thus furnish hydrogen ions. Example: HCl

Iron(II) sulfide reacts with nonoxidizing acid, say hydrochloric acid to give rotten egg smelling gas, H₂S (Hydrogen sulfide).

The reaction is shown below:

FeS (s) + 2HCl (aq) ⇒ FeCl₂ (s) + H₂S (g)

A general reaction is also shown below of the reaction of Iron(II) sulfide with any nonoxidizing acid, HX as:

FeS (s) + 2HX (aq) ⇒ FeX₂ (s) + H₂S (g)

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A sample of gas with an initial volume of 12.5 L at a pressure of 784 torr and a temperature of 295 K is compressed to a volume
Kipish [7]

Answer:

Final pressure in (atm) (P1) = 6.642 atm

Explanation:

Given:

Initial volume of gas (V) = 12.5 L

Pressure (P) = 784 torr

Temperature (T) = 295 K

Final volume (V1) = 2.04 L

Final temperature (T1) = 310 K

Find:

Final pressure in (atm) (P1) = ?

Computation:

According to combine gas law method:

\frac{PV}{T} =\frac{P1V1}{T1} \\\\\frac{(784)(12.5)}{295} =\frac{(P1)(2.04)}{310}\\\\33.22 = \frac{(P1)(2.04)}{310}\\\\P1=5,048.18877

⇒ Final pressure (P1) = 5,048.18877 torr

⇒ Final pressure in (atm) (P1) = 5,048.18877 torr / 760

⇒ Final pressure in (atm) (P1) = 6.642 atm

3 0
3 years ago
Why do the planets keep orbiting the sun
dsp73
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3 0
3 years ago
You need to determine the specific gravity of a sample. After putting the sample on a lab scale, you know it has a mass of 85 gr
BlackZzzverrR [31]

Answer:

Specific gravity of the sample = 8.947

Explanation:

Specific gravity of a substance is defined as the density of that substance divided by the density of water.

Density of water = 1000g/l

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= 85/9.5 x 10^-3

= 8947.37 g/l

SG = 8947.37/1000

= 8.947

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3 years ago
Where is a divergent boundary most likely to be found?
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Answer:

a

Explanation

Bc I took the test, good luck with that guys

7 0
2 years ago
if 100. mL of 0.800 M Na2SO4 is added to 200. mL of 1.20 M NaCl, what is the concentration of Na+ ions in the final solution? As
Black_prince [1.1K]
Compounds Na₂SO₄ and NaCl are mixed together are we are asked to find the concentration of Na⁺ in the mixture 
Na₂SO₄ ---> 2 Na⁺ + SO₄³⁻

1 mol of Na₂SO₄ gives out 2 mol of Na⁺ ions 

the number of Na₂SO₄ moles added - 0.800 M/1000 * 100 ml
                                                         = 0.08 mol
therefore number of Na⁺ ions from Na₂SO₄ = 0.08 * 2 = 0.16 mol

NaCl ----> Na⁺ + Cl⁻ 
1 mol of NaCl gives 1 mol of Na⁺ ions
number of NaCl moles added = 1.20 M/1000 * 200 ml
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number of Na⁺ ions from NaCl = 0.24 mol

total number of Na⁺ ions in the mixture = 0.16 mol + 0.24 mol = 0.4 mol
as stated the volumes are additive, 
therefore total volume  = 100 ml + 200 ml = 300 ml 
the concentration of Na⁺ ions = number of moles / volume 
                                              = 0.4 mol/ 0.3 dm³
concentration of Na⁺ = 1.33 mol/dm³
5 0
3 years ago
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