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Anuta_ua [19.1K]
2 years ago
15

Help i know whofkfkfkf

Chemistry
1 answer:
Irina18 [472]2 years ago
7 0

<u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>:</u><u>-</u>

1. A compound contains atoms of different elements chemically combined together in a fixed ratio. An element is a pure chemical substance made of same type of atom. ... Elements that compose the compound are chemically combined. Elements contain only one type of atom.

2. Two ways to add energy to break down a compound are to apply heat and to apply an electric current.

3. The compound composed of these ions exhibits properties entirely different from the properties of the elements sodium and chlorine. Chlorine is poisonous, but sodium chloride is essential to life; sodium atoms react vigorously with water, but sodium chloride simply dissolves in water.

4. In chemistry, the law of definite proportion, sometimes called Proust's law, or law of constant composition states that 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.

5. compounds and elements.

6.Solid nonmetals

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which two human body system function together to distribute oxygen for use in cellular process throughout the body
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3 years ago
Suppose a student titrates a 10.00-ml aliquot of saturated ca(oh)2 solution to the equivalence point with 16.08 ml of 0.0199 m h
Alborosie
Following chemical reaction is involved upon titration of Ca(OH)2 with HCl,
Ca(OH)2 + 2HCl ↔ CaCL2 + 2H2O

Above is an example of acid-base titration to generate salt and water. Here, H+ ions of acid (HCl) combines with OH- (ions) of base [Ca(OH)2] to generated H2O

Given,
concentration of HCl = 0.0199 M
Total volume of HCl consumed during titration = 16.08 mL = 16.08 X 10^(-3) L

∴, number of moles of H+ consumed = Molarity X Vol. of HCl (in L)
                                                           = 0.0199 X 16.08 X 10^(-3)
                                                           = 3.1999 X 10^-4 mol
Thus, total number of moles of [OH-] ions present initial = 3.1999 X 10-4 mol
So, initial conc. [OH-] ion = 
\frac{number of moles of [OH-]}{volume of solution (L)} = \frac{3.1999 X 10^(^-^4^)}{10 X 10^(^-^3^)} = 0.03199 M
6 0
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How many moles of H, O can be formed from 3.02 x 1023 atoms of NH from the
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Hey, I don't know but try 10+ (h) + 5 ), 79

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Which statement is true?
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A I think

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