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Fofino [41]
3 years ago
13

20 POINTS! what would be an example of an everyday household acid and base?

Chemistry
2 answers:
Elza [17]3 years ago
4 0

Answer:

Baking Soda, Vinegar, Ammonia

Explanation:

Bogdan [553]3 years ago
4 0

vinegar is a household acid and washing soda is a very strong base

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Chromium (Cr) can combine with Chlorine (Cl2) to form Chromium chloride (CrCl3). Which equation is a correction representation o
Shalnov [3]
Balance the reaction so the same number of each type of atom is on each side. Which option seems balanced to you?
4 0
2 years ago
Read 2 more answers
I will give brainliest. If you burn the carbon in limited air, the reaction is
Fynjy0 [20]

This reaction is different in that the carbon undergoes an incomplete combustion as opposed to complete combustion where carbon is fully oxidized. A caveat: incomplete combustion products in general can be difficult to predict without sufficient information, as it's not uncommon to obtain a mixture of different products.

Here, we are told that solid carbon is burned in limited air to produce a gas. I am presuming that, in the equation that's given, the "0" represents a blank where you must fill in a chemical symbol. In this case, our equation would be: 2C(s) + O₂(g) → 2CO(g).

There is not enough information here to provide the numerical answers to the two questions. From the words in the question (e.g., "is different" and "this time"), it would seem that this question is an excerpt from a larger or preceding question where specific numbers had been provided or computed.

However, it's possible to make some general observations on how one may go about answering these questions <em>if </em>one had more information.

Since we're to assume that oxygen is the limiting reagent, if one is given the amount of solid carbon (either in mass, moles, or number of atoms), it's possible to determine the moles of CO(g) that's produced since C and CO have an equal stoichiometric ratio. So, for example, if one burns 2 moles of C(s), then 2 moles of CO(g) would be produced.

<em><u>But</u></em>, there is still not enough information to compute the volume of CO gas if this is the line of questioning. We don't know, for instance, the temperature or pressure of the reaction conditions. In fact, the only way it would be possible to answer this would be if you were given beforehand a conversion factor that relates the volume of CO(g) to its quantity (e.g., to assume that one mole of gas occupies <em>x </em>liters).

As for the second question, this would depend on what you know about the quantity of the C(s) reacted and/or the quantity (or volume, from question a) of CO(g) produced. If you can get the number of moles of C(s) reacted or CO(g) produced, the number of moles of O₂(g) used up: It would be half the number of moles of C(s) reacted or half the number of moles of CO(g) produced). <u>Again</u>, it's impossible to determine the volume of O₂(g) using just the information provided here, so I suspect that you must have further information relating gas quantity to volume. As we did with CO(g), the volume of O₂(g) used up can be found using whatever conversion factor you have.

If you have any further information or questions, please feel free to follow up.  

6 0
2 years ago
For the chemical equation SO 2 ( g ) + NO 2 ( g ) − ⇀ ↽ − SO 3 ( g ) + NO ( g ) the equilibrium constant at a certain temperatur
MakcuM [25]

Answer:

Moles of NO₂ = 0.158

Explanation:

                         SO 2 ( g ) + NO 2 ( g ) ⇄  SO 3 ( g ) + NO ( g )

              According to the law of mass equation

                                     

                                       K_{c} = \frac{[SO_{3} ][NO]}{[SO_{2}][NO_{2}  ]}

                              ⇒   3.10 = \frac{(1.00)(1.00)}{(2.30) [NO_{2} ]}    At equilibrium [SO₃] = [NO]

                              ⇒ [NO₂] = \frac{1}{6.3}

                              ⇒ [NO₂] = 0.158

So. number of moles of NO₂ at equilibrium added = 0.158

8 0
3 years ago
vanillin is a flavoring agent in vanilla it has a mass percent composition of 61.15% c,5.3%H,31.55% o determine the empirical fo
Nutka1998 [239]

The empirical and molecular formulas will be  C_5H_5O_2 and  C_1_0H_1_0O_4 respectively.

<h3>Empirical and molecular formula</h3>

The compound contains C, H, and O.

C = 61.15/12 = 5.0958

H = 5.3/1 = 5.3

O = 31.55/16 = 1.9719

Divide by the smallest

C = 2.6

H = 2.7

O = 1

Thus, the empirical formula is C_5H_5O_2

Empirical formula mass = (12x5) + (1x5) + 16x2 = 97

n = 152.15/97 = 2

The molecular formula is C_1_0H_1_0O_4

More on molecular and empirical formulas can be found here: brainly.com/question/14425592

#SPJ1

6 0
2 years ago
What class of organism is vital in the recycling of nutrients in the ecosystem?
asambeis [7]
I believe that would be a decomposer
6 0
3 years ago
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