Answer:
47 is your correct ans, which is b
Answer:
<em>Because the individual components of any mixture are not</em> <u>bonded</u> t<em>o each other, the composition of those components can vary. Also, some of the </em><u>physical</u> <em>properties of the individual components are still noticeable.</em>
Explanation:
A <em>mixture</em> is a combination of two or more pure substances that are present in any proportion and each pure substance keeps its own physical and chemical properties.
As oppossite to mixtures, the compounds are pure substances formed by two or more different elements which are chemically bonded to each other. So, while in the compounds the components (elements) are bonded in a fixed proportion, and their composition cannot vary, in the mixtures each component may be present in any proportion, which means that the <em>composition can vary</em>.
Take, for example, the simple case of talc and iron particles.This is <em>a mixture</em>. Talc is <u><em>not bonded</em></u> to the iron particles, and so their proportion, <em>the compositoin</em>, can vary in any form. Aslo, both talc and iron particles keep their own <u><em>physical properties</em></u>: you can perfectly separate the mixture by using a magnet to attract the iron particles, because they have not lost their magnetic property (a physical one).
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Answer:
15.95
Explanation:
This question is a modification of the calculation of the empirical formula of a compound given its percent composition and atomic weights of the elements in the compound.
Here we are given the formula and the percent composition, so we know that there are 4 atoms of E per 2 atoms of N so lets solve using the information given.
In 100 grams of the binary compound we have
30.46 g N
69.54 g E
The number of moles is the mass divided by atomic weight:
mol N = 30.46 g / A.W N = 30.46 g / 14.00 g/mol = 2.18 mol N
mol E = 65.54 g / A.W E
Thus,
4 mol E/ 2 mol N = ( 69.54 g/ A.W E ) / 2.18
2 A.E = 65.54 g / 2.18 ⇒ A.W E = 69.54 g / ( 2 x 2.18 ) = 15.94 g
So the A.W is 15.94 g/mol which is close the atomic weight of O.
The equivalency point is at the point of the titration where the amount of titrant added neutralize the solution. When it’s a strong acid strong base titration, the equivalence point will be 7. When it is a weak acid strong base, the equivalence point it more basic (the exact number depends on what acid and base you use). And when it is a strong acid weak base, the equivalence number is more acid (the exact number depends on what acid and base you use). Hope this helps!
