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mr Goodwill [35]
3 years ago
7

Calculate the total volume of gas (at 119 ?c and 731 mmhg produced by the complete decomposition of 1.71 kg of ammonium nitrate.

Chemistry
1 answer:
Shtirlitz [24]3 years ago
4 0
The reaction is:

NH4 (NO3) (s) ⇄ N2O (g) + 2 H2O (g)

This means that 1 mol of NH4 (NO3)s produces 3 moles of gases.

Now find the number of moles in 1.71 kg of NH4 (NO3)

Molar mass = 2*14g/mol + 4 * 1g/mol + 3*16g/mol = 80 g/mol

# moles = mass / molar mass = 1710 g / 80 g/mol = 21.375 mol of NH4(NO3)

We already said that every mol of NH4(NO3) produces 3 moles of gases, then the number of moles of gases produced is 3 * 21.375 = 64.125 mol

Now use the equation for ideal gases to fin the volume

pV = nRT => V = nRT / p = (64.125 mol)(0.082atm*liter / K*mol) * (119 +273)K / (731mmHg *1 atm/760mmHg) =

V = 2143.01 liters
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A student constructs a coffee cup calorimeter and places 50.0 mL of water into it. After a brief period of stabilization, the te
Grace [21]

Answer:

The calorimeter constant is  = 447 J/°C

Explanation:

The heat absorbed or released (Q) by water can be calculated with the following expression:

Q = c × m × ΔT

where,

c is the specific heat

m is the mass

ΔT is the change in temperature

The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.

The heat absorbed by the calorimeter (Q) can be calculated with the following expression:

Q = C × ΔT

where,

C is the calorimeter constant

The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).

Qabs + Qrel = 0

Qabs = - Qrel

Qcal + Qw₁ = - Qw₂

Qcal = - (Qw₂ + Qw₁)

Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)

Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) +  (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]

Ccal  = 447 J/°C

5 0
3 years ago
2. 27.8 mL of an unknown were added to a 50.0-mL flask that weighs 464.7 g. The total mass of the flask and the liquid is 552.4
agasfer [191]

Answer:

d=4.24x10^{-4}\frac{lb}{in^3}

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to set the equation for the calculation of density and mass divided by volume:

d=\frac{m}{V}

Thus, we can find the mass of the unknown by subtracting the total mass of the liquid to the mass of the flask and the liquid:

m=552.4g-464.7g=87.7g

So that we are now able to calculate the density in g/mL first:

d=\frac{87.7g}{27.8mL}=3.15g/mL

Now, we proceed to the conversion to lb/in³ by using the following setup:

d=3.15\frac{g}{mL}*\frac{1lb}{453.6g}*\frac{1in^3}{16.3871mL}\\\\d=4.24x10^{-4}\frac{lb}{in^3}

Regards!

6 0
3 years ago
Does data from mass spectrometry indicate that modern scientists have made modifications to Dalton''s model? Justify.
Gelneren [198K]
John Dalton was a scientist who proposed that all matter consists of atoms. At this stage, no one had yet discovered neutrons and the nucleus. As a result, Dalton's model consisted of a single atom i.e. the atom was the smallest object. 
A mass spectrometer is an instrument that is able to see what is inside an atom. Scientists have been able to prove that the item is not the smallest object in the world. Atoms are made up of smaller objects called protons, neutrons and electrons. 
We can, therefore, safely conclude that data  from mass spectrometry has helped modern scientists to make modifications to Dalton's model. <span>
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5 0
3 years ago
Helppppppppppppppppppppppppppp
Charra [1.4K]

Kinetic energy remains conserved in an elastic collision.

5 0
2 years ago
A 35.40 gram hydrate of sodium carbonate, Na2CO3•nH2O, is heated to a constant mass. Its final weight is 30.2 g. What is formula
Sever21 [200]

Answer:

Na₂CO₃•H₂O

Explanation:

After it is heated, the remaining mass is the mass of sodium carbonate.

30.2 g Na₂CO₃

Mass is conserved, so the difference is the mass of the water:

35.4 g − 30.2 g = 5.2 g H₂O

Convert masses to moles:

30.2 g Na₂CO₃ × (1 mol Na₂CO₃ / 106 g Na₂CO₃) = 0.285 mol Na₂CO₃

5.2 g H₂O × (1 mol H₂O / 18.0 g H₂O) = 0.289 mol H₂O

Normalize by dividing by the smallest:

0.285 / 0.285 = 1.00 mol Na₂CO₃

0.289 / 0.285 = 1.01 mol H₂O

The ratio is approximately 1:1.  So the formula of the hydrate is Na₂CO₃•H₂O.

3 0
3 years ago
Read 2 more answers
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