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max2010maxim [7]

2 years ago

12

You have a two-liter sample of each of the following gases, all at 25 °C and one atmosphere pressure. Which sample will weigh th

e most?A) Ammonia.B) Argon.C) Carbon dioxide.D) Nitrogen dioxide.E) Oxygen.

1 answer:

stepladder [879]2 years ago

4 0

Answer: D) Nitrogen Dioxide

Explanation:

You might be interested in

State the oxidation number of S in <br><img src="https://tex.z-dn.net/?f=H_%7B2%7DSO_%7B3%7D" id="TexFormula1" title="H_{2}SO_{3

Nezavi [6.7K]

Taking into account the definition of oxidation number, the oxidation numbers of S in H₂SO₃ is 4.

<h3>Definition of oxidation number</h3>

The oxidation number is the charge that an atom has; is an integer that represents the number of electrons an atom puts into play when it forms a given compound.

In other words, the oxidation number of an element is a value that indicates the number of electrons that element gains or loses when it combines with another.

<h3>Oxidation number determination</h3>

To determine the oxidation state of different elements it is necessary to know that:

The oxidation number of hydrogen in a compound is +1, except in metal hydrides, where is –1.
The oxidation number of oxygen in a compound is –2, except in peroxides, where it is –1.

On the other side, the sum of the oxidation numbers of the existing elements in a chemical formula must add up to zero.

Then, considering the oxidation numbers of each element, multiplying it by the number of existing elements in the chemical formula and adding it and equaling it to zero, the value of the missing oxidation number can be obtained.

<h3>Oxidation numbers of S</h3>

In this case, the oxidation numbers of S in H₂SO₃ is calculated as:

2× (+1) + oxidation numbers of S + 3×(-2)= 0

2 + oxidation numbers of S -6= 0

oxidation numbers of S -4= 0

<u><em>oxidation numbers of S= 4</em></u>

Finally, the oxidation numbers of S in H₂SO₃ is 4.

Learn more about the oxidation number:

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5 0

2 years ago

Read 2 more answers

The heat of fusion for water is 80. cal/g. How many calories of heat are needed to melt a 35 g ice cube that has a temperature o

HACTEHA [7]

Answer: 2800 calories

Explanation:

Latent heat of fusion is the amount of heat required to convert 1 mole of solid to liquid at atmospheric pressure.

Amount of heat required to fuse 1 gram of water = 80 cal

Mass of ice given = 35 gram

Heat required to fuse 1 g of ice at $0^0C$ = 80 cal

Thus Heat required to fuse 35 g of ice = $\frac{80}{1}\times 35=2800cal$

Thus 2800 calories of energy is required to melt 35 g ice cube

6 0

3 years ago

An amphoteric species is neither an acid nor a base. True or False

umka2103 [35]

Answer: False

Explanation: just took on edge

8 0

2 years ago

sp2 hybrid orbitals have 1. trigonal bipyramidal symmetry. 2. linear symmetry. 3. tetrahedral symmetry. 4. trigonal planar symme

Cloud [144]

Answer:

4. trigonal planar symmetry.

Explanation:

The sp2 hybridization is formed with one s and two p atomic orbitals and form trigonal planar symmetry. In sp2 hybridization, there are same valence shell in both the orbitals and it gives three equivalent sp2 hybridized orbitals that are separated by 120 degrees giving trigonal planar symmetry.

Hence, the correct answer is "4. trigonal planar symmetry.".

7 0

2 years ago

A gas used to extinguish fires is composed of 75 % CO2 and 25 % N2. It is stored in a 5 m3 tank at 300 kPa and 25 °C. What is th

tatyana61 [14]

Answer : The partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

Explanation :

As we are given 75 % $CO_2$ and 25 % $N_2$ in terms of volume.

First we have to calculate the moles of $CO_2$ and $N_2$ .

$\text{Moles of }CO_2=\frac{\text{Volume of }CO_2}{\text{Volume at STP}}=\frac{75}{22.4}=3.35mole$

$\text{Moles of }N_2=\frac{\text{Volume of }N_2}{\text{Volume at STP}}=\frac{25}{22.4}=1.12mole$

Now we have to calculate the mole fraction of $CO_2$ .

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CO_2+\text{Moles of }N_2}$

$\text{Mole fraction of }CO_2=\frac{3.35}{3.35+1.12}=0.75$

Now we have to calculate the partial pressure of the $CO_2$ gas.

$\text{Partial pressure of }CO_2=\text{Mole fraction of }CO_2\times \text{Total pressure of gas}$

$\text{Partial pressure of }CO_2=0.75mole\times 300Kpa=225Kpa=225Kpa\times \frac{0.145\text{ psia}}{1Kpa}=32.625\text{ psia}$

conversion used : (1 Kpa = 0.145 psia)

Therefore, the partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

3 0

2 years ago