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In-s [12.5K]
3 years ago
13

Which of the following dienes would you expect to be the most stable? CH3CH2CH=CHCH2CH=CHCH3 CH3CH=CHCH=CHCH2CH3 CH2=CHCH2CH2CH2

CH=CH2 CH2=CHCH=CHCH2CH2CH3 CH3CH=C(CH3)CH=CHCH2CH3
Chemistry
1 answer:
saveliy_v [14]3 years ago
4 0

Answer:

CH3CH=C(CH3)CH=CHCH2CH3

Explanation:

This question is all about determining the stability for Dienes. Dienes are divided into three major parts that is the cumulated diene, Isolated dienes, and conjugated dienes.

Among the three types or classes of dienes, the conjugated type of diene is the most stable. That is to say that among the five compounds given in the question the most stable is CH3CH=C(CH3)CH=CHCH2CH3 which is a conjugated diene.

The reason behind the stability of this compound, that is CH3CH=C(CH3)CH=CHCH2CH3 is because of what is known as delocalization . Hybridization and Resonance are what causes the delocalization which involves pie interactions with alpha hydrogen, also, another reason behind the stability is the lower heat of hydrogenation.

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The chemical formula for copper (II) phosphate is Cu3(PO4)2. What is the charge on each copper ion?
mrs_skeptik [129]
3+ because I just know it 8757890
6 0
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So, a gas, and an
valina [46]

Answer:

d) cut the large sized Cu solid into smaller sized pieces

Explanation:

The aim of the question is to select the right condition for that would increases the rate of the reaction.

a) use a large sized piece of the solid Cu

This option is wrong. Reducing the surface area decreases the reaction rate.

b) lower the initial temperature below 25 °C for the liquid reactant, HNO3

Hugher temperatures leads to faster reactions hence this option is wrong.

c) use a 0.5 M HNO3 instead of 2.0 M HNO3

Higher concentration leads to increased rate of reaction. Hence this option is wrong.

d) cut the large sized Cu solid into smaller sized pieces

This leads to an increased surface area of the reactants, which leads to an increased rate of the reaction. This is the correct option.

5 0
3 years ago
The molar mass is determined by measuring the freezing point depression of an aqueous solution. A freezing point of -5.20°C is r
Dima020 [189]

Answer:

The empirical formula is C2H4O3

The molecular formula is C4H8O6

The molar mass is 152 g/mol

Explanation:

The complete question is: An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Step 1: Data given

Mass % of Carbon = 31.57 %

Mass % of H = 5.30 %

Freezing point = -5.20 °C

10.56 grams of the compound dissolved in 25.0 grams of water

Kf water = 1.86 °C kg/mol

Step 2: Calculate moles of Carbon

Suppose 31.57% = 31.57 grams

moles C = mass C / Molar mass C

moles C = 31.57 grams / 12.0 g/mol = 2.63 moles

Step 3: Calculate moles of Hydrogen:

Moles H = 5.30 grams / 1.01 g/mol

moles H = 5.25 moles

Step 4: Calculate moles of Oxygen

Moles O = ( 100 - 31.57 - 5.30) / 16 g/mol

Moles O = 3.95 moles

Step 5: We divide by the smallest number of moles

C: 2.63 / 2.63 = 1 → 2

H: 5.25/2.63 = 2 → 4

O: 3.95/ 2.63 = 1.5 → 3

The empirical formula is C2H4O3

The molar mass of the empirical formula = 76 g/mol

Step 6: Calculate moles solute

Freezing point depression = 5.20 °C = m * 1.86

m = 5.20 / 1.86

m = 2.80 molal = 2.80 moles / kg

2.80 molal * 0.025 kg = 0.07 moles

Step 7: Calculate molar mass

Molar mass = mass / moles

Molar mass = 10.56 grams / 0.07 moles

Molar mass = 151 g/mol

Step 8: Calculate molecular formula

151 / 76 ≈  2

We have to multiply the empirical formula by 2

2*(C2H4O3) = C4H8O6

The molecular formula is C4H8O6

The molar mass is 152 g/mol

6 0
3 years ago
The density of water at 40°c is 0.992 g/ml. what is the volume of 3.45 g of water at this temperature?
wariber [46]
3.45÷0.992=3.48ml
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8 0
3 years ago
Calculate the molality of a 20.0% by mass ammonium sulfate (nh4)2so4 solution. the density of the solution is 1.117 g/ml.
olasank [31]
Hello!

We have the following data:

m1 (solute mass) = 20 % m/m
M1 (Molar mass of solute) (NH4)2 SO4 = ?
m2 (mass of the solvent) = ? (in Kg)

First we find the solute mass (m1), knowing that:

20% m/m = 20g/100mL

20 ------ 100 mL (0,1 L)
y g --------------- 1 L

y = 20/0,1 
y = 200 g --> m1 = 200 g

Let's find Solute's Molar Mass, let's see:

M1 of (Nh4)2SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

We must find the volume of the solvent and therefore its mass (m2), let us see:

d = 1,117 g/mL
m = 200 g
v (volumen of solute) = ?

d =  \dfrac{m}{V} \to V =  \dfrac{m}{d}

V =  \dfrac{200\:\diagup\!\!\!\!g}{1,117\:\diagup\!\!\!\!g/mL} \to V = 179\:mL\:(volumen\:of\:solute)

<span>The solvent volume will be:
</span>
1000 -179 => V = 821 mL (volumen of disolvent)

If: 1 mL = 1g

<span>Then the mass of the solvent is:
</span>
m2 (mass of the solvent) = 821 g → m2 (mass of the solvent) = 0,821 Kg

Now, we apply all the data found to the formula of Molality, let us see:

\omega =  \dfrac{m_1}{M_1*m_2}

\omega =  \dfrac{200}{132*0,821}

\omega =  \dfrac{200}{108,372}

\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark

_________________________________
_________________________________


<span>Another way to find the answer:
</span>
We have the following data: 

W (molality) = ? (in molal)
n (number of mols) = ?
m1 (solute mass) = 20 % m/m = 20g/100mL → (in g to 1L) = 200 g
m2 (disolvent mass) the remaining percentage, in the case: 80 % m/m = 800 g → m2 (disolvent mass) = 0,8 Kg
M1 (Molar mass of solute) (NH4)2 SO4 
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol 


<span>Let's find the number of mols (n), let's see:

</span>n =  \dfrac{m_1}{M_1}

n = \dfrac{200}{132}

n \approx 1,5\:mol

Now, we apply all the data found to the formula of Molality, let us see:

\omega =  \dfrac{n}{m_2}

\omega =  \dfrac{1,5}{0,8}

&#10;\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark

I hope this helps. =)
7 0
3 years ago
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