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pickupchik [31]
3 years ago
8

The molecules of a substance formed a rigid repeating crystal structure. Then, a phase change occurred. That structure is now mo

re like clusters of molecules moving past one another. What phase change caused this change? Plz, explain your answer.
A
condensing
B
evaporating
C
freezing
D
melting
Chemistry
1 answer:
BartSMP [9]3 years ago
6 0

Answer: everybody report fNnasks he doesn't answer nobody question if you see it like you told me do it yourself science is super easy

Explanation:

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Passing an electric current through a sample of water (H2O) can cause the water to decompose into hydrogen gas (H2) and oxygen g
Andreas93 [3]
2H2O --> 2H2 + O2
The mole H2O:mole O2 ratio is 2:1
Now determine how many moles of O2 are in 50g: 50g × 1mol/32g = 1.56 moles O2
Since 1 mole of O2 was produced for every 2 moles of H2O, we need 2×O2moles = H2O moles
2×1.56 = 3.13 moles H2O
Finally, convert moles to grams for H2O:
3.13moles × 18g/mol = 56.28 g H2O
D) 56.28
7 0
4 years ago
The molar solubilities of the following compounds (in mol/L) are:
IceJOKER [234]

<u>Answer:</u> The decreasing order of K_{sp} is AgSCN>AgBr>AgCN

<u>Explanation:</u>

  • <u>For AgBr:</u>

The balanced equilibrium reaction for the ionization of silver bromide follows:

AgBr\rightleftharpoons Ag^{+}+Br^-

                 s      s

The expression for solubility constant for this reaction will be:

K_{sp}=[Ag^{+}][Br^-]

We are given:

s=7.3\times 10^{-7}mol/L

Putting values in above equation, we get:

K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.3\times 10{-7})^2=5.33\times 10^{-13}

Solubility product of AgBr = 5.33\times 10^{-13}

  • <u>For AgCN:</u>

The balanced equilibrium reaction for the ionization of silver cyanide follows:

AgCN\rightleftharpoons Ag^{+}+CN^-

                    s      s

The expression for solubility constant for this reaction will be:

K_{sp}=[Ag^{+}][CN^-]

We are given:

s=7.7\times 10^{-9}mol/L

Putting values in above equation, we get:

K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(7.7\times 10{-9})^2=5.93\times 10^{-17}

Solubility product of AgCN = 5.33\times 10^{-17}

  • <u>For AgSCN:</u>

The balanced equilibrium reaction for the ionization of silver thiocyanate follows:

AgSCN\rightleftharpoons Ag^{+}+SCN^-

                     s      s

The expression for solubility constant for this reaction will be:

K_{sp}=[Ag^{+}][SCN^-]

We are given:

s=1.0\times 10^{-6}mol/L

Putting values in above equation, we get:

K_{sp}=s\times s\\\\K_{sp}=s^2\\\\K_{sp}=(1.0\times 10{-6})^2=1.0\times 10^{-12}

Solubility product of AgSCN = 1.0\times 10^{-12}

The decreasing order of K_{sp} follows:

AgSCN>AgBr>AgCN

4 0
4 years ago
Can someone help with this question plsss!!
Nadusha1986 [10]

Answer:

d

Explanation:

6 0
3 years ago
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Which molecule is shown below?
antoniya [11.8K]

Answer:

D. 3-methylheptane

Explanation:

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3 years ago
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Q#5. The bond between which two atoms is most polar? ​
Ede4ka [16]

Answer:

H-O

Explanation:

because oxygen and hydrogen have very high difference in electronegativity

8 0
3 years ago
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