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3 years ago

How many resonance structures are there for [S2CO]2- ?

Chemistry

1 answer:

Irina-Kira [14]3 years ago

5 0

Answer:

Three

Explanation:

1. Draw a Lewis structure

Put the least electronegative atom (C) in the centre
Attach the O and two S atoms to the C.
Add enough electrons to give each atom an octet.

2. Draw the resonance structures

You should get one of the structures below.

It has a double bond and two single bonds.

The resonance structures will have the double bond in each of the other two positions.

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P is the pressure in atmospheres (atm), V is the volume in liters (L), n is the number of moles, R is the gas constant (0.0821 L

yKpoI14uk [10]

Answer:

The mass of the neon gas m = 1.214 kg

Explanation:

Pressure = 3 atm = 304 k pa

Volume = 0.57 L = 0.00057 $m^{3}$

Temperature = 75 °c = 348 K

Universal gas constant = 0.0821 $\frac{L . atm}{mol K}$

We have to change the unit of this constant. it may be written as

Universal gas constant = 8.314 $\frac{KJ}{mol K}$

Gas constant for neon = $\frac{8.314}{20}$ = 0.41 $\frac{KJ}{kg K}$

From ideal gas equation,

P V = m R T ------- (1)

We have all the variables except m. so we have to solve this equation for mass (m).

⇒ 304 × $10^{3}$ × 0.00057 = m × 0.41 × 348

⇒ 173.28 = 142.68 × m

⇒ m = 1.214 kg

This is the mass of the neon gas.

4 0

3 years ago

calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o

Leviafan [203]

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of $Na_3PO_4$ = 12.00 g

Mass of $CaCl_2$ = 10.0 g

Molar mass of $Na_3PO_4$ = 164 g/mol

Molar mass of $CaCl_2$ = 111 g/mol

Molar mass of $NaCl$ = 58.5 g/mol

Molar mass of $Ca_3(PO_4)_2$ = 310 g/mol

First we have to calculate the moles of $Na_3PO_4$ and $CaCl_2$ .

$\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}$

$\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol$

and,

$\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}$

$\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

$2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2$

From the balanced reaction we conclude that

As, 3 mole of $CaCl_2$ react with 2 mole of $Na_3PO_4$

So, 0.0901 moles of $CaCl_2$ react with $\frac{2}{3}\times 0.0901=0.0601$ moles of $Na_3PO_4$

From this we conclude that, $Na_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $CaCl_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$ and $Ca_3(PO_4)_2$

From the reaction, we conclude that

As, 3 mole of $CaCl_2$ react to give 6 mole of $NaCl$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{6}{3}\times 0.0901=0.1802$ mole of $NaCl$

and,

As, 3 mole of $CaCl_2$ react to give 1 mole of $Ca_3(PO_4)_2$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{1}{3}\times 0.0901=0.030$ mole of $Ca_3(PO_4)_2$

Now we have to calculate the mass of $NaCl$ and $Ca_3(PO_4)_2$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

$\text{ Mass of }NaCl=(0.1802moles)\times (58.5g/mole)=10.5g$

and,

$\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2\times \text{ Molar mass of }Ca_3(PO_4)_2$

$\text{ Mass of }Ca_3(PO_4)_2=(0.030moles)\times (310g/mole)=9.3g$

Therefore, the mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

5 0

3 years ago

Katyanochek1 [597]

C because a compound is a substance made of at least two atoms bonded together

4 0

3 years ago

What's the IUPAC name pf this compound? CH3-CH-CH2-CH-CHO | | CH3 CL

Svetradugi [14.3K]

Answer:

2-chloro-4-methylpentanal.

Explanation:

Hello there!

In this case, according to the chemical compound:

CH3-CH-CH2-CH-CHO

| |

CH3 Cl

We can see the main functional group is an starting carbonyl, which means this is an aldehyde. Moreover, we can see a Cl-substituent on the second carbon and a methyl substituent on the fourth carbon. Therefore, the IUPAC name turns out: 2-chloro-4-methylpentanal.

Best regards!

8 0

3 years ago

A chemist dilutes 185 of the 3.0 M solution with water until a new volume of 750 is obtained. Calculate the new concentration

svlad2 [7]

Answer:

0.74M

Explanation:

Step 1 :

Data obtained from the question.

Initial concentration (C1) = 3M

Initial volume (V2) = 185mL

Final volume (V2) = 750mL

Final concentration (C2) =..?

Step 2:

Determination of the new concentration of the solution.

The new concentration of the solution can be obtained by using the dilution formula as shown below:

C1V1 = C2V2

3 x 185 = C2 x 750

Divide both side by 750

C2 = 3 x 185 / 750

C2 = 0.74M

Therefore, the new concentration of the solution is 0.74M

7 0

3 years ago