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Leni [432]
3 years ago
8

Ammonium nitrite undergoes decomposition to produce only gases as shown below. nh4no2(s) → n2(g) + 2h2o(g) how many liters of ga

s will be produced by the decomposition of 32.0 g of nh4no2 at 525°c and 1.5 atm?
Chemistry
1 answer:
schepotkina [342]3 years ago
3 0
From the given balanced reaction equation:

we can see that every 1 mole of NH4NO2 will produce 1 mole of N2 and when the water also on the gas phase so, every 1 mole of NH4NO2 will produce 3 moles of gas

now we need to get number of moles of NH4NO2  = mass/molar mass
when the mass = 32 g and the molar mass of NH4NO2 = 64 mol/g

∴moles of NH4NO2 = 32 g / 64 mol/g = 0.5 mole

so, the moles that produced of gas will be = 3 * 0.5 = 1.5 moles

now, we can use the ideal gas formula to get V:

PV = nRT

when P is the pressure = 1.5 atm

n is the number of moles = 1.5 moles 

R is the ideal gas constant = 0.0821 L*atm/mol*K

and T is the temperature in Kelvin = 525 °C + 273 = 798 K

so by substitution:

1.5 atm * V = 1.5 moles * 0.0821 * 798 K

∴ V = 65.5 L
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konstantin123 [22]

Answer:

T°fussion of solution is -18°C

Explanation:

We have to involve two colligative properties to solve this. Let's imagine that the solute is non electrolytic, so i = 1

First of all, we apply boiling point elevation

ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of pure solvent

Kb =  ebuliloscopic constant

105°C - 100° = 0.512 °C kg/mol  . m . 1

5°C / 0.512 °C mol/kg = m

9.7 mol/kg = m

Now that we have the molality we can apply, the Freezing point depression.

ΔT = Kf . m . i

Kf =  cryoscopic constant

0° - (T°fussion of solution) = 1.86 °C/m  . 9.76 m . 1

- (1.86°C /m . 9.7 m) = T°fussion of solution

- 18°C = T°fussion of solution

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3 years ago
One way in which elements differ from each other is the structure of the electron cloud in each elements Atoms in electron cloud
BartSMP [9]

Not sure what you are asking. I have two possible answers though...

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Again, I don't know what you were asking, but one of these answers may be correct.

4 0
3 years ago
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3 years ago
Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates. The phosphate ion forms in
katrin2010 [14]

Answer:

22.4269 grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions

Explanation:

We will first write the balanced equation for this scenario

3 CaCl2 + 2 Na3PO4 ----> 6 NaCl + Ca3 (PO4)2

3 Mg(NO3)2 + 2 Na3PO4 -----> 6 NaNO3 + Mg3 (PO4)2

The ratio here for both calcium chloride and magnesium nitrate is 3:2

The number of moles of each compound is equal to

0.054 * 1.4 = 0.0756\\0.093* 1.4 = 0.1302

Using the mole ratio of 3:2, convert each to moles of sodium phosphate.

0.0756 mole of CaCl2 is equal to 0.05\\ Na3PO4

0.1302 mole of CaCl2 is equal to 0.0868 Na3PO4

Converting moles of sodium phosphate to grams of sodium phosphate we get

(0.05 +0.0868) * 163.94 g/mol

22.4269 grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions

8 0
3 years ago
WHAT IS THE PERCENT BY VOLUME OF ETHANOL IN A SOLUTION THAT CONTAINS 35 mL ETHANOL IN 115 mL OF WATER?
Dmitry [639]
We need to first add both of the solution volumes together 35+115=150. Now we can divide the volume of the ethanol by the total volume 35/150=.233. To double check we can multiply the total volume by the percentage of ethanol by volume we got as a solution 150x.233=35. So the percentage by volume of ethanol in the solution is .233x100=23.3%.
3 0
3 years ago
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