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OLEGan [10]
3 years ago
12

PLEASE HURRY. Which of the molecules below is propyne?

Chemistry
2 answers:
Oduvanchick [21]3 years ago
7 0

The last photo because its molecular formula is C3H4.

Valentin [98]3 years ago
6 0

Answer: Option (D) is the correct answer.

Explanation:

General formula of an alkyne is C_{n}H_{2n-2}.

Suffix "yne" in the name of a compound represents that there is a triple bond present in the compound.

So, a molecule of propyne will have three carbon atoms, four hydrogen atoms and a triple bond. Therefore, there wll be 6 sigma bonds and 2 pi bonds in a propyne molecule.

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When a voltaic cell operates, ions move through a salt bridge. The presence of the salt bridge completes the circuit because it promotes ion flow. It maintains the charges balance since the electron are moving from a half cell to another.
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3 years ago
This homework assignemnt is confusing can someone please help.
Juliette [100K]

Answer: Heyya, the answer is  0.8539gH2. Pls mark brainliest

Explanation:

5 0
2 years ago
What is the approximate value of the ccc bond angle in ch3ch2ch2oh
Fynjy0 [20]

Answer:

The value of the carbon bond angles are 109.5 °

Explanation:

CH3CH2CH2OH = propanol . This is an alcohol.

All bonds here are single bonds.

Single bonds are sp³- hybdridization type. To be sp3 hybridized, it has an s orbital and three p orbitals : sp³. This refers to the mixing character of one 2s-orbital and three 2p-orbitals. This will  create four hybrid orbitals with similar characteristics.

Sp3- types have angles of 109.5 ° between the carbon - atoms.

This means that the value of the carbon bond angles are 109.5 °

6 0
4 years ago
Calculate the molecular mass of the element
Elanso [62]

Answer:

Cu3 (PO4)2

3×64 + 2×(31+ 4×16)

192 + 2×(31+64)

192 + 2×(95)

192+190

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4 0
2 years ago
The weak acid HA is 4 % ionized (dissociated) in a 0.30 M solution. A. What is Ka for this acid?B. What is the pH of this soluti
ANTONII [103]

Answer:

Ka=5x10^{-4}

pH=1.92

Explanation:

Hello,

In this case, given the percent ionization and the concentration of the acid, one computes the concentration of hydrogen ions as follows:

\% ionization=\frac{[H^+]}{[HA]}*100\%

[H^+]=\frac{\% ionization*[HA]}{100\%} =\frac{4\%*0.30M}{100\%}=0.012M

Therefore the Ka is computed by using the equilibrium expression:

Ka=\frac{[H^+][A^-]}{[HA]} =\frac{0.012M*0.012M}{0.30M-0.012M}\\ \\Ka=5x10^{-4}

And the pH:

pH=-log([H^+])=-log(0.012)\\\\pH=1.92

Regards.

4 0
3 years ago
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