Given :
Mass of oxygen containing carbon monoxide (CO) is 2.666 gram .
To Find :
How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O) .
Solution :
By law of constant composition , a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.
So , volume of solution does not matter .
Moles of oxygen , 
.
Now , molecule of CO contains 1 mole of C .
So , moles of C is also 0.167 mole .
Mass of carbon , 
.
Therefore , mass of carbon is 2 grams .
Hence , this is the required solution .
the correct IUPAC name of the compound is 1-Butanal.
<h3>What are IUPAC names?</h3>
It is a system of naming organic compounds based on the longest carbon-to-carbon single bonds. It does not matter whether these longest chains are continuous or in a ring.
Thus, when the compound with the chemical formula, CH3-CH2-CH2CHO is considered. The longest carbon-to-carbon chain is 4. The 1st carbon carries a functional group known as an aldehyde.
Aldehydes are equipped with the carbonyl group and have the general formula R−CH=O. They are also sometimes referred to as formyl.
Aldehydes are named after their parent alkane chains with a slight modification. The 'e' is replaced with 'al'
The aldehyde in this case has four carbons. This means that the parent alkane is Butane. Therefore, the name of the compound will be 1-Butanal.
More on IUPAC names can be found here: brainly.com/question/16631447
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Answer:
Average atomic mass = 17.5 amu.
Explanation:
Given data:
X-17 isotope = atomic mass17.2 amu, abundance:78.99%
X-18isotope = atomic mass 18.1 amu, abundance 10.00%
X-19isotope = atomic mass:19.1 amu, abundance: 11.01%
Average atomic mass of X = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass = (78.99×17.2)+(10.00×18.1) +(11.01+ 19.1) /100
Average atomic mass = 1358.628 + 181 +210.291 / 100
Average atomic mass = 1749.919 / 100
Average atomic mass = 17.5 amu.
Answer:
A. Rutherford and Bohr
Explanation:
I think this is right but I am not sure it has been a while since i read about them
It is very important<span> to know the shape of a molecule if one is to understand its reactions. It is also desirable to have a simple method to predict the geometries of compounds. For main group compounds, the </span>VSEPR<span> method is such a predictive tool and unsurpassed as a handy predictive method.</span>
