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KonstantinChe [14]
2 years ago
5

Which of the following is NOT

Chemistry
1 answer:
Ronch [10]2 years ago
5 0

Answer:

The answer is neither

Explanation:

All of the following things all have atoms in them therefor,its neither.

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What type of bond forms between two oxygen atoms
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Answer:A double convalescent bond is where two pairs of electrons are shared between the atoms rather than just one pair. Two oxygen atoms can both achieve stable structures by sharing two pairs of electrons as in the diagram.

Explanation:

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Radio waves are ____________ waves​
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Answer: an electromagnetic wave of a frequency between about 104 and 1011 or 1012 Hz, as used for long-distance communication.

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3 years ago
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What is the number of moles of chemical units represented by 9.03x10^24? And how do I show work? (DUMB IT DOWN PLEASE)
Goryan [66]
You have to use avagados number... so toy take your starting number and multiple by 1/avagados number which is like 6.002^23 and that will equal your number of males... avagado always associates with moles... look up that exact number though bc I cant remember it
6 0
2 years ago
Methylamine (ch3nh2) is a weakly basic compound. calculate the kb for methylamine if a 0.253 m solution is 4.07% ionized.
MariettaO [177]
Answer is: Kb for methylamine is 4.37·10⁻⁴.<span>
Chemical reaction: CH</span>₃NH₂ + H₂O → CH₃NH₃⁺ + OH⁻.
c(CH₃NH₂) = 0.253 M.
α = 4.07% ÷ 100% = 0.0407.
[CH₃NH₃⁺] = [OH⁻] = c(CH₃NH₂) · α.
[CH₃NH₃⁺] = [OH⁻] = 0.253 M · 0.0407.
[CH₃NH₃⁺] = [OH⁻] = 0.0103 M.
[CH₃NH₂] = 0.253 M - 0.0103 M.
[CH₃NH₂] = 0.2427 M.
Kb = [CH₃NH₃⁺] · [OH⁻] / [CH₃NH₂]. 
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Kb = 4.37·10⁻⁴.
8 0
3 years ago
Determine the pH of 0.050 M HCN solution. HCN is a weak acid with a Ka equal to 4.9 x 10-10<br> DONE
nadezda [96]

Answer:

The pH of the solution is 5.31.

Explanation:

Let "\alpha is the dissociation of weak acid - HCN.

The dissociation reaction of HCN is as follows.

                  HCN+H_{2}O\rightarrow H_{3}O^{+}+CN^{-}

Initial                  C                         0            0

Equilibrium        c(1- \alpha)              c\alpha c\alpha

Dissociation constant = Ka= c\alpha \times \frac{c\alpha}{c(1-\alpha)}

=\frac{c\alpha^{2}}{(1-\alpha)}

In this case weak acids \alpha is very small so, (1-\alpha ) is taken as 1.

Ka=C\alpha^{2}

\alpha=\sqrt\frac{ka}{c}

From the given the concentration = 0.050 M

Substitute the given value.

\alpha=\sqrt\frac{4.9\times 10^{-10}}{0.05}=9.8\times 10^{-4}

[H_{3}O^{+}]=c\alpha

[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}

pH= -log[H_{3}O^{+}]

=-log[4.9\times10^{-6}]

=6-log 4.9= 5.31

Therefore, The pH of the solution is 5.31.

7 0
3 years ago
Read 2 more answers
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