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Kisachek [45]
3 years ago
12

How much H2O is produced when 20000 g of C2H2 burns completely? Answer in units of g.

Chemistry
1 answer:
IgorC [24]3 years ago
3 0

Answer:

Explanation:

H  = 1

C = 12

O = 16

Acetylene, HC≡CH = 2+24 = 26

H2O = 2 + 16 = 18

In XS oxygen, one HC≡CH yields one H2O

26 g HC≡CH  ==> 18 g H2O

2000 g HC≡CH ==> 2000*18/26 g H2O =  1384.6154 g H2O

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We are given with the conditions of the weather balloon and required to determine the volume of the balloon when the altitude is raised to 25,000 feet. We determine the number of moles through ideal gas law PV = nRT. Plugging to this equation, the number of moles is 1.1351 moles. We substitute this with another set of conditions, volume is 47.46 liters. 
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3 years ago
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In your own words, describe what is the role or function of the neutron in the atom's nucleus
pochemuha

Neutrons are very important in provide stability for an atom. When atoms are created by fusion, neutrons are included in this process. Because protons don't like each other and repel each other that's where neutrons come in.

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3 years ago
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An "empty" container is not really empty if it contains air. How may moles of nitrogen are in an "empty" two-liter cola bottle a
Lisa [10]

Answer:

1. 0.0637 moles of nitrogen.

2. The partial pressure of oxygen is 0.21 atm.  

Explanation:

1. If we assume ideal behaviour, we can use the Law of ideal gases to find the moles of nitrogen, considering that air composition is mainly nitrogen (78%), oxygen (21%) and argon (1%):  

V_{N_2}=V_{T}\times 0.78=2L \times 0.78 =1.56 L\\PV=nRT\\n_{N_2}=\frac{PV}{RT}=\frac{1 atm\times 1.56 L}{0.0821\frac{atmL}{molK}\times 298 K}\\n_{N_2}= 0.0637 mol

2. Now, in order to find he partial pressure of oxygen we need to find the total moles of air, and then the moles of oxygen. Then, we use these results to determine the molar fraction of oxygen, to multiply it with total pressure and get the partial pressure of oxygen as follows:

n_{total}=\frac{1 atm \times 2L}{0.0821 \frac{atmL}{molK}298K}=0.0817 mol

V_{O_2}=2L \times 0.21 = 0.42 L\\n_{O_2}=\frac {1atm \times 0.42 L}{0.0821 \frac{atm L}{mol K}298 K}=0.0172 mol\\X_{O_2}=\frac{n_{O_2}}{n_{total}}=\frac{0.0172 mol}{0.0817 mol}= 0.21

P_{O_2}=X_{O_2} \times P = 1 atm \times 0.21 = 0.21 atm

As you see, the molar fraction and volume fraction are the same because of the assumption of ideal behaviour.  

3 0
3 years ago
Taylor has a mass or 500 grams
aliya0001 [1]

Answer:

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Explanation:

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8 0
2 years ago
g Sucrose (C12H22O11), a nonionic solute, dissolves in water (normal freezing/melting point 0.0°C) to form a solution. If some u
DIA [1.3K]

Answer:

15.4 g of sucrose

Explanation:

Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m

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

Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m

0.56°C / 1.86 m/°C = m → 0.301 mol/kg

m → molality (moles of solute in 1kg of solvent)

Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg

0.301 mol/kg .  0.150kg = 0.045 moles.

We determine the mass of sucrose, by the molar mass:

0.045 mol . 342 g/1mol = 15.4 g

4 0
3 years ago
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