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rewona [7]
2 years ago
3

Name the term used for the configuration of a stereoisomer having R groups whose positions alternate from one side to the other

side along the backbone chain.
Chemistry
1 answer:
Naddik [55]2 years ago
5 0

Answer:

Syndiotactic

Explanation:

Syndiotactic is the term used for the configuration of a stereoisomer having R groups whose positions alternate from one side to the other side along the backbone chain.

In simple words , they are polymers that have the substitute group arranged in alternating manner.

For example, Syndiotic polystyrene,  Gutta percha  etc.

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What mass of methanol is combusted in a reaction that produces 112 L of Co2 at STP?
AlekseyPX

Answer

D 160g

Explanation:

<u>Write the equation:</u>

Combustion reactions use oxygen and release water and heat, so

  CH₃OH(g) + O₂(g) → CO₂(g) + H₂O(g)

Balance that:

  2CH₃OH(g) + 3O₂(g) → 2CO₂(g) + 4H₂O(g)

<u>Find moles of carbon dioxide:</u>

We need to know the number of moles of CO₂. This rxn is at STP, so at STP one mole of gas = 22.4 liters.

  112 L * 1 mol/22.4 L = <em>5 mol CO₂</em>

<u>Find moles of methanol:</u>

Based on the chemical equation, for every 2 mol methanol, there are 2 mol carbon dioxide. So for every 5 mol carbon dioxide, there are 5 mol methanol!

  5 mol CO₂ = 5 mol CH₃OH

Molar mass of methanol: 12.01 + 3*1.008 + 16.00 + 1.008 = <em>32.04 g/mol</em>

Moles of methanol: 5 mol * 32.04 g/mol = 160.2 g methanol

≈ 160 mol methanol

8 0
2 years ago
Jesse travels 3.0 meters east and then turns and travels 4.0 meters north. The trip requires 35 seconds. What is his velocity? 0
melisa1 [442]
Displacement = √(3² + 4²)
Displacement = 5 meters north east

Velocity = displacement / time
Velocity = 5 / 35
Velocity = 0.14 m/s northeast
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Read 2 more answers
How many grams o an ointment base must be added to 45 g o clobetasol (T EMOVAT E) ointment, 0.05% w/w, to change its strength to
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Answer:

Drug calculation

If we have 45g of clobetasol  = 0.05%w/w

Then what mass in g of clobetasol is in 0.03%w/w = 45 x 0.03/0.05 =27g

It means that 27g of clobetasol must be added to change the drug strength to 0.03% w/w

3 0
2 years ago
Which conjugate pair is suited best to make this buffer? Which conjugate pair is suited best to make this buffer? Phosphoric aci
Juliette [100K]

Answer:

bkfhjjfrsxtr

Explanation:

4 0
2 years ago
One mole of an ideal gas in a closed system, initially at 25°C and 10 bar, is first expanded adiabatically, then heated isochori
Igoryamba

Answer:

P_2=0.398bar=39800Pa

T_2=118.7K\\

Q=-3729.9J

W=-61753.24J

ΔU_T=0J

ΔH_T=0J

Explanation:

Hello,

At the first state, the molar volume is:

v_1=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^6Pa}=2.48x10^{-3}m^3

The volume in both the second and third state:

v_2=v_3=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^5Pa}=2.48x10^{-2}m^3

Now, as it is about an adiabatic process, one remembers the following relationships:

PV^\alpha =K\\TV^{\alpha-1}\\\alpha=\frac{Cp}{Cv}=\frac{7/2R}{5/2R}=1.4

- Next, for the aforesaid volumes and the first pressure, one computes the second pressure as:

P_2=\frac{P_1V_1^\alpha }{V_2^\alpha} =\frac{10bar*(2.48x10^{-3}m^3)^{1.4}}{(2.48x10^{-2}m^3)^{1.4}} =0.398bar=39800Pa

- And the temperature:

T_2=\frac{T_1V_1^{\alpha-1}}{V_2^{\alpha-1}} =\frac{298.15K*(2.48x10^{-3}m^3)^{1.4-1}}{(2.48x10^{-2}m^3)^{1.4-1}} =118.7K\\

- Q:

It is clear that the heat for the first process is 0 as it is adiabatic, but for the second one, it is computed as:

Q_2=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J

Then the total heat:

Q=Q_1+Q_2=0-3729.9J=-3729.9J

- The work for the first process is:

W_1=\frac{P_2V_2-P_1V_1}{1-\alpha }=\frac{39800Pa*2.48x10^{-3}m^3-1x10^6Pa*2.48x10^{-2}m^3}{0.4} \\W_1=-61753.24J

It is clear that the second process is isochoric, so the work here is zero, thus, the total work is:

W=W_1+W_2=-61753.24J+0J=-61753.24J

- For the two processes, ΔU becomes the same value since the system returns to the initial temperature, so ΔU total is 0, thus, for each process, one's got:

U_1=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J\\U_2=nCv(T_3-T_2)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=3729.9J\\

- Finally, the total enthapy is also 0 due to same aforesaid reason, thus, each enthalpy is:

H_1=nCp(T_2-T_1)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-5221.86J\\H_2=nCv(T_3-T_2)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=5221.86J\\

Best regards.

8 0
2 years ago
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