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Troyanec [42]
3 years ago
7

How many grams of water can be produced when 68 L of oxygen gas react

Chemistry
2 answers:
Cloud [144]3 years ago
5 0
2H2 + O2 -> 2H20 is the balanced equation for the reaction of Hydrogen with Oxygen to form water so If you have 32g of O2 this is a simple Dimensional analysis problem 32g O2 x 36.03056g H20/31.9988g O2 this way the O2 cancels out and you are left with just the H2O so your raw answer would be 36.0319112, then if your instructor requires a significant figure answer that would be to 2 significant figures the information you were given 32g O2, so as above 36g or Water are produced. Just a different way to view and solve the problem with the balanced equation so you can see the way everything relates to everything else. the molar masses of O2 and H2O are simply found my adding up 2 Oxygens 15.9994g x2 = 31.9988g and H2O = 2(1.00794) + 15.9994 = 18.01258 but you then have to multiply that by 2 because the reaction states you get 2 mols in the reaction so that is where I came up with the 36.03056g for the solution. Hope this helps. Dimensional Analysis is and can be one of the best ways to solve these problems, because not always are you going to be dealing with 1:2 ratios. 
WARRIOR [948]3 years ago
3 0
If you have 32g of O2 this is a simple Dimensional analysis problem 32g O2 x 36.03056g H20/31.9988g O2 this way the O2 cancels out and you are left with just the H2O so your raw answer would be 36.0319112, then if your instructor requires a significant figure answer that would be to 2 significant figures the information you were given 32g O2, so as above 36g or Water are produced. Just a different way to view and solve the problem with the balanced equation so you can see the way everything relates to everything else. the molar masses of O2 and H2O are simply found my adding up 2 Oxygens 15.9994g x2 = 31.9988g and H2O = 2(1.00794) + 15.9994 = 18.01258 but you then have to multiply that by 2 because the reaction states you get 2 mols in the reaction so that is where I came up with the 36.03056g for the solution. Hope this helps. Dimensional Analysis is and can be one of the best ways to solve these problems, because not always are you going to be dealing with 1:2 ratios. 
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Below are the steps of the engineering process (not necessarily in order): 1) Know the background 2) Do the work 3) Make a plan
LuckyWell [14K]

Answer:

4 - 1 - 3 - 2 - 6 - 5

Explanation:

During an engineering process, first, we need to identify the problem, or the need because the process only will occur because of some need. Then, it's necessary to know as much as possible about the problem and the things that already exist or already were tested to solve it. Knowing the background will make the work easy.

After that, it's necessary to plan the things we'll do, knowing the costs, the time needed for activities, how many people will be necessary for each step, etc. It's really important to make a plan. Then, do the work, following the plan. Thus, the process must be tested. During the test of the results, some problems must be found, so it's time to evaluate and redesign the process, to solve these problems found.

3 0
3 years ago
1.2 L sample of gas is determined to contain 0.5 moles of nitrogen. At the same temperature and pressure, how many moles of gas
hram777 [196]

A 20 L sample of the gas contains 8.3 mol N₂.

According to <em>Avogadro’s Law,</em> if <em>p</em> and <em>T</em> are constant

<em>V</em>₂/<em>V</em>₁ = <em>n</em>₂/<em>n</em>₁

<em>n</em>₂ = <em>n</em>₁ × <em>V</em>₂/<em>V</em>₁

___________

<em>n</em>₁ = 0.5 mol; <em>V</em>₁ = 1.2 L

<em>n</em>₂ = ?;           <em>V</em>₂ = 20 L

∴<em>n</em>₂ = 0.5 mol × (20 L/1.2 L) = 8.3 mol

4 0
3 years ago
Write the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each:
Maru [420]

Answer :

(a) The charge and full ground-state electron configuration of the monatomic ion is, (+1) and 1s^22s^22p^63s^23p^64s^23d^{10}4p^6

(b) The charge and full ground-state electron configuration of the monatomic ion is, (-3) and 1s^22s^22p^6

(c) The charge and full ground-state electron configuration of the monatomic ion is, (-1) and 1s^22s^22p^63s^23p^64s^23d^{10}4p^6

Explanation :

For the neutral atom, the number of protons and electrons are equal. But, they are unequal when the atoms present in the form of ions or the atom has some charges.

When an unequal number of electrons and protons then it leads to the formation of ionic species.

Ion : An ion is formed when an atom looses or gains electron.

When an atom looses electrons, it will form a positive ion known as cation.

When an atom gains electrons, it will form a negative ion known as anion.

(a) The given element is, Rb (Rubidium)

As we know that the rubidium element belongs to group 1 and the atomic number is, 37

The ground-state electron configuration of Rb is:

1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^1

This element will easily loose 1 electron and form Rb^+ ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Rb ion is:

1s^22s^22p^63s^23p^64s^23d^{10}4p^6

(b) The given element is, N (Nitrogen)

As we know that the nitrogen element belongs to group 15 and the atomic number is, 7

The ground-state electron configuration of N is:

1s^22s^22p^3

This element will easily gain 3 electrons and form N^{3-} ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of N ion is:

1s^22s^22p^6

(c) The given element is, Br (Bromine)

As we know that the bromine element belongs to group 17 and the atomic number is, 35

The ground-state electron configuration of Rb is:

1s^22s^22p^63s^23p^64s^23d^{10}4p^5

This element will easily gain 1 electron and form Br^- ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Br ion is:

1s^22s^22p^63s^23p^64s^23d^{10}4p^6

4 0
3 years ago
Which of these form an ionic bond
zhenek [66]

Answer:

the second 1

Explanation:

7 0
3 years ago
Why hyberdization occurs?
Damm [24]
Because the resulting hybridized orbitals are more stable
8 0
3 years ago
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