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3 years ago

NEED HELP ASAP!!! I just need to know the order

Chemistry

1 answer:

sweet-ann [11.9K]3 years ago

7 0

4 Movement of less dense material
3 Heating of cooler material
1 cooling of warmer material
2 movement of denser material

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If I have 0.070 moles of gas at a pressure of 0.20 atm and at a temperature of 8.00°C,

Ivanshal [37]

Answer:

PV=nRT

0.20×v = 0.070×8.00

0.20V= 0.56

0.20v÷0.20v = 0.56÷0.20v

= 2.8

3 0

3 years ago

Propose a mechanism to account for the formation of a cyclic acetal from 4-hydroxypentanal and one equivalent of methanol. If th

timofeeve [1]

Answer:

If carbonyl oxygen of 4-hydroxypentanal is enriched with $O^{18}$ , then the oxygen label appears in the water .

Explanation:

In the first step, -OH group at C-4 gives intramolecular nucleophilic addition reaction at carbonyl center to produce a cyclic hemiacetal.
Then, one equivalent of methanol gives nucleophilic substitiution reaction by substituting -OH group in cyclic hemiacetal to produce cyclic acetal.
If carbonyl oxygen of 4-hydroxypentanal is enriched with $O^{18}$ , then the oxygen label appears in the water produced at the end of reaction.
Full reaction mechanism has been shown below.

8 0

4 years ago

The standard reduction potentials for the Ag+|Ag(s) and Zn2+| Zn(s) half-cell reactions are +0.799 V and -0.762 V, respectively.

mihalych1998 [28]

Answer: The potential of the given cell is 1.551 V

Explanation:

The given chemical cell follows:

$Zn(s)|Zn^{2+}(0.125M)||Ag^{+}(0.240M)|Ag(s)$

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(0.125M)+2e^-;E^o_{Zn^{2+}/Zn}=-0.762V$

Reduction half reaction: $Ag^{+}(0.240M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.799V$ ( × 2)

Net cell reaction: $Zn(s)+2Ag^{+}(0.240M)\rightarrow Zn^{2+}(0.125M)+2Ag(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.799-(-0.762)=1.561V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Ag^{+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = ? V

$E^o_{cell}$ = standard electrode potential of the cell = +1.561 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=0.125M$

$[Ag^{+}]=0.240M$

Putting values in above equation, we get:

$E_{cell}=1.561-\frac{0.059}{2}\times \log(\frac{(0.125)}{(0.240)^2})$

$E_{cell}=1.551V$

Hence, the potential of the given cell is 1.551 V

6 0

4 years ago

Consider the following data on some weak acids and weak bases

SpyIntel [72]

Answer:

b < c < a < d

Explanation:

The weak acid with the lowest pKa will be the most acidic. In the other way, the conjugate base which the acid is weak will be strong.

The weak base with the lowest pKb will be the most basic. And the conjugate base of the weak base will be a strong acid.

Ka Acetic acid = 1.8x10-5

Ka HCN = 1.9x10-10

Kb pyridine = 1.7x10-9

Kb NH3 = 1.8x10-5

NH4Br is the conjugate base of a weak base. That means is a strong acid.

NH4Br has the lowest pH

NaBr is the conjugate base of a strong acid, HBr. That means NaBr is neutral

The most basic between the conjugate base of the acetic acid, NaCH3CO2 and KCN is KCN because the acetic acid is the stronger acid regard to HCN.

The rank is:

NH4Br < NaBr < NaCH3CO2 < KCN

6 0

3 years ago

The volume of a sample of gas is 0.5 L and the pressure is 0.98 atm. The volume is increased to 1.0 L. Show the set

meriva

Given,

P1 = 0.98 atm

V1 = 0.5 L

V2 = 1.0 L

P2 = ?

Solution,

According to Boyle's Law,

P1V1 = P2V2

0.98 × 0.5 = 1.0 × P2

P2 = 0.98 × 0.5 × 1.0

P2 = 0.49 atm

Answer - The new pressure is 0.49 atm.

5 0

3 years ago