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ki77a [65]
3 years ago
9

9.710 x 10-8 grams of Chy are how many molecules? Use Cl = 35.45 g/mole.

Chemistry
1 answer:
aliya0001 [1]3 years ago
6 0

Answer:

26.42

Explanation:

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Name each subatomic particle, its charge, and its location in an atom.
Tpy6a [65]
Protons: charge +1, have a mass of 1 and are found in the nucleus 

Neutrons: charge 0, have a mass of 1 and are found in the nucleus

Electrons: charge -1, have a mass of 1/840 and are found on the outside of the nucleus 

hope that helps 
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3 years ago
The four elements that make up about 96% of body matter are ________.
AveGali [126]

An element is a pure substance that cannot be separated or divided into simpler substances by any method (physical or chemical). The four common elements in all the living organisms are carbon, oxygen, hydrogen, nitrogen.

Thus, the four elements that make up about 96 % of body matter are carbon, oxygen, hydrogen, nitrogen (c).

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Which of the following metals is in a liquid state at room temperature?
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3 years ago
What is the energy conversion that occurs in cellular respiration?
fomenos
Cellular respiration is the process by which the chemicalenergy of "food" molecules is released and partially captured in the form of ATP. Carbohydrates, fats, and proteins can all be used as fuels in cellular respiration, but glucose is most commonly used as an example to examine the reactions and pathways involved.
3 0
3 years ago
The maximum amount of nickel(II) cyanide that will dissolve in a 0.220 M nickel(II) nitrate solution is...?
sweet [91]

Answer : The maximum amount of nickel(II) cyanide is 5.84\times 10^{-12}M

Explanation :

The solubility equilibrium reaction will be:

                       Ni(CN)_2\rightleftharpoons Ni^{2+}+2CN^-

Initial conc.                        0.220       0

At eqm.                             (0.220+s)   2s

The expression for solubility constant for this reaction will be,

K_{sp}=[Ni^{2+}][CN^-]^2

Now put all the given values in this expression, we get:

3.0\times 10^{-23}=(0.220+s)\times (2s)^2

s=5.84\times 10^{-12}M

Therefore, the maximum amount of nickel(II) cyanide is 5.84\times 10^{-12}M

7 0
3 years ago
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