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3 years ago

How many sigma and pi bonds are present in the molecule hcn?

Chemistry

1 answer:

Leni [432]3 years ago

8 0

Answer:

Hydrogen cyanide (HCN) contains two (2) sigma bonds and two (2) pi bonds.

Explanation:

Hydrogen cyanide can be represented structurally as :

H-C≡ N

From the structural formula, we can see:

One C-H sigma bond

While the C≡ N bond contains One Sigma and two (2) pi bonds.

You might be interested in

How does an indicator differentiate between an acid and base?

Svet_ta [14]

Answer:

that is all I know sorry hope that may help you :)

Explanation:

Indicator is a substance which shows different colors in acidic and basic medium. Indicators can be natural (derived from natural sources) or artificial ( man-made) for example :

*Litmus is an indicator. Acid turns blue litmus into red while base turns red litmus into blue

* Turmeric solution does not show change in color (remains yellow) in acidic solution and turns red in basic solution.

8 0

2 years ago

3. How many moles of NaCl are present in 6000. La 1.5 M NaCl solution?

Leno4ka [110]

Answer:

Option A (9.0) is the correct alternative.

Explanation:

The given values are:

Molarity,

= 1.5 M

Volume,

= 6000 mL

or,

= 6 L

As we know,

⇒ $Molarity=\frac{Moles}{Volume}$

or,

⇒ $The \ moles \ of \ NaCl=Molarity\times Volume$

By putting the values, we get

$=1.5\times 6$

$=9 \ mol$

7 0

3 years ago

in a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M merc

Margarita [4]

Answer: 0.1161 grams of mercury(II) sulfide) form.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$ .....(1)

a) Molarity of $Na_2S$ solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

$0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol$

$\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol$

b) Molarity of $Hg(NO_3)_2$ solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

$0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol$

$Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3$

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ reacts with 1 mole of $Na_2S$

Thus 0.0005 moles of $HgNO_3$ reacts with= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg(NO_3)_2$

Thus $Hg(NO_3)_2$ is the limiting reagent and $Na_2S$ is the excess reagent.

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ forms= 1 mole of $Hg_2S$

Thus 0.0005 moles of $Hg(NO_3)_2$ forms= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg_2S$

mass of $H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g$

Thus 0.1161 grams of mercury(II) sulfide) form.

5 0

3 years ago

C. If 62.9 g of lead (II) chloride is produced, how many grams of lead (II) nitrate were

melomori [17]

Answer:

Mass = 76.176 g

Explanation:

Given data:

Mass of lead(II) chloride produced = 62.9 g

Mass of lead(II) nitrate used = ?

Solution:

Chemical equation:

Pb(NO₃)₂ + 2HCl → PbCl₂ + 2HNO₃

Number of moles of lead(II) chloride:

Number of moles = mass/molar mass

Number of moles = 62.9 g/ 278.1 g/mol

Number of moles = 0.23 mol

Now we will compare the moles of lead(II) chloride with Pb(NO₃)₂ from balance chemical equation:

PbCl₂ : Pb(NO₃)₂

1 : 1

0.23 : 0.23

Mass of Pb(NO₃)₂:

Mass = number of moles × molar mass

Mass = 0.23 mol × 331.2 g/mol

Mass = 76.176 g

8 0

4 years ago

What happen in the dark reaction

Firdavs [7]

Answer

Dark reactions make use of these organic energy molecules (ATP and NADPH). This reaction cycle is also called Calvin Benison Cycle, and it occurs in the stroma. ATP provides the energy while NADPH provides the electrons required to fix the CO2 (carbon dioxide) into carbohydrates.

5 0

3 years ago