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3 years ago

Which is an example of a three

Chemistry

2 answers:

malfutka [58]3 years ago

6 0

Answer:

It's A. 120 cubic meters

Minchanka [31]3 years ago

4 0

It’s a.120 cubic meters

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Methane reacts with oxygen to produce carbon dioxide and water,What are the products? A. methane and water. B. oxygen and carbon

EleoNora [17]

D, carbon dioxide and water. Methane and oxygen PRODUCE carbon dioxide and water in their reaction.

3 0

4 years ago

Read 2 more answers

Why does a solid have a definite shape and volume?

stellarik [79]

Answer:

Explanation:

Particles in a solid have fixed locations in a volume that does not change. Solids have a definite volume and shape because particles in a solid vibrate around fixed locations.

6 0

3 years ago

Sulfur and oxygen form both sulfur dioxide and sulfur trioxide. When samples of these were decomposed the sulfur dioxide produce

Zinaida [17]

Answer : The mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.

Explanation : Given,

Mass of oxygen in sulfur dioxide = 3.49 g

Mass of sulfur in sulfur dioxide = 3.50 g

Mass of oxygen in sulfur trioxide = 9.00 g

Mass of sulfur in sulfur trioxide = 6.00 g

Now we have to calculate the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide.

Mass of oxygen per gram of sulfur for sulfur dioxide = $\frac{\text{Mass of oxygen}}{\text{Mass of sulfur}}$

Mass of oxygen per gram of sulfur for sulfur dioxide = $\frac{3.49}{3.50}=0.997g$

and,

Mass of oxygen per gram of sulfur for sulfur trioxide = $\frac{\text{Mass of oxygen}}{\text{Mass of sulfur}}$

Mass of oxygen per gram of sulfur for sulfur trioxide = $\frac{9.00}{6.00}=1.5g$

Thus, the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.

8 0

3 years ago

What is the formula for tetrabromine decafluoride

Lyrx [107]

Answer:

$Br_{4}F_{10}$

The prefix "Tetra" implies 4 Bromine atoms. The prefix "Deca" implies 10 fluorine atoms.

4 0

3 years ago

How many hours does it take to form 15.0 L of O₂ measured at 750 torr and 30°C from water by passing 3.55 A of current through a

rosijanka [135]

Answer:

The correct answer is 17.845 hours.

Explanation:

To solve the question, that is, to determine the hours required there is a need to combine the Faraday's law of electrolysis with the Ideal gas law.

Based on Faraday's law, m = Mit/nF

Here m is the mass in grams, M is the molecular mass, i is the current in amperes, t is time, n is the number of moles of electron per mole of oxygen formed and F is the Faraday's constant (the value of F is 96487 coulombs/mole).

From the above mentioned equation,

t = mnF/Mi ------(i)

Now based on ideal gas law's, PV = nRT or PV = m/M RT, here n = mass/molecular mass.

So, from the above gas law's equation, m = PVM/RT

Now putting the values of m in the equation (i) we get,

t = PVMnF/MiRT = PVnF/iRT

Based on the given information, the value of P is 750 torr or 750/760 atm = 0.98 atm, the value of v is 15.0 L, T is 30 degree C or 273 + 30 K = 303 K, i is 3.55 Amperes, and the value of R is 0.0821 atm L/mol K.

1 mole of oxygen gives 2 moles of electrons, therefore, 2 moles of oxygen will give 4 moles of electrons.

Now putting the values we get,

t = PVnF/iRT

= 0.98 atm × 15.0 L × 4 moles of electron × 96487 coulombs per mole / 3.55 coulomb per sec × 0.0821 atm L per mole-K × 303 K

= 64243.81 secs or 64243.81/3600 hr

= 17.845 hours

7 0

3 years ago