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RUDIKE [14]
4 years ago
12

Describe the possible components of a buffer solution. (select all that apply.)strong base and its conjugate acidweak base and i

ts conjugate acid
Chemistry
1 answer:
Viktor [21]4 years ago
6 0
Missing question:

a) strong base and its conjugate acid.

b) weak base and its conjugate acid.

c) strong acid and strong base.

d) weak acid and its conjugate base.

e) weak acid only.

f) weak base only.

g) strong acid and its conjugate base.

Answers are: 

b) weak base and its conjugate acid. For example weak base ammonia (NH₃) and its conjugate acid ammonium ion (NH₄⁺).

d) weak acid and its conjugate base. For example weak acetic acid (CH₃COOH) and its conjugate base CH₃COO⁻.

Buffer solution changes pH very little when a small amount of strong acid or base<span> is added to solution.</span>


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When a field is declared static, there will be:
Arturiano [62]

Answer:

B. only one copy of the field in memory

Explanation:

A static method is sort of a description of a class but is not part of the objects that it generates. Crucial: A program may perform a static method without constructing an object first! All other functions (those not static) only occur when they're member of an object. Thus it is necessary to build an object before they could be executed.

Therefore, when an static field is declared static, there will be:

B. only one copy of the field in memory

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3 years ago
What is the difference between luster and dull
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<h2>Explanation:</h2><h2> </h2>

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3 years ago
A cube of butter weighs .254 pounds and has a volume of 129.4 milliliters what is the density
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d = .254/129.4
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Read 2 more answers
Given the following information regarding the masses and relative abundances for the isotopes of an element, determine its atomi
Ilia_Sergeevich [38]

Answer:

Average atomic mass = 58.51047 amu

The symbol is Ni

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})+(\frac {\%\ of\ the\ third\ isotope}{100}\times {Mass\ of\ the\ third\ isotope})+(\frac {\%\ of\ the\ fourth\ isotope}{100}\times {Mass\ of\ the\ fourth\ isotope})+(\frac {\%\ of\ the\ fifth\ isotope}{100}\times {Mass\ of\ the\ fifth\ isotope})

Given that:

For first isotope:

% = 68.274 %

Mass = 57.9353 amu

For second isotope:

% = 26.095 %

Mass = 58.9332 amu

For third isotope:

% = 1.134 %

Mass = 61.9283 amu

For fourth isotope:

% = 3.593 %

Mass = 63.9280 amu

For fifth isotope:

% = 0.904 %

Mass = 63.9280 amu

Thus,  

Average\ atomic\ mass=\frac{68.274}{100}\times {57.9353}+\frac{26.095}{100}\times {58.9332}+\frac{1.134}{100}\times {61.9283}+\frac{3.593}{100}\times {63.9280}+\frac{0.904}{100}\times {63.9280}

Average\ atomic\ mass=39.55474 +15.37861854+0.702266922+2.29693304+0.57790912

Average atomic mass = 58.51047 amu

The symbol is Ni

5 0
4 years ago
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