Can someone pls answer quickly?

Qual substância apresenta 2 (dois) pares de elétrons sendo compartilhados entre seus átomos na

-Dominant- [34]

Answer:

Explanation:

a bucusev 35ft

3 0

3 years ago

What can be said about an Endothermic reaction with a negative entropy change?The reaction isa. spontaneous at all temperatures.

Lyrx [107]

Answer:

1. d. The reaction is spontaneous in the reverse direction at all temperatures.

2. c. The reaction is spontaneous at low temperatures.

Explanation:

The spontaneity of a reaction is associated with the Gibbs free energy (ΔG). When ΔG < 0, the reaction is spontaneous. When ΔG > 0, the reaction is non-spontaneous. ΔG is related to the enthalpy (ΔH) and the entropy (ΔS) through the following expression:

ΔG = ΔH - T. ΔS [1]

where,

T is the absolute temperature (T is always positive)

<em>1. What can be said about an Endothermic reaction with a negative entropy change?</em>

If the reaction is endothermic, ΔH > 0. Let's consider ΔS < 0. According to eq. [1], ΔG is always positive. The reaction is not spontaneous in the forward direction at any temperature. This means that the reaction is spontaneous in the reverse direction at all temperatures.

<em>2. What can be said about an Exothermic reaction with a negative entropy change?</em>

If the reaction is exothermic, ΔH < 0. Let's consider ΔS < 0. According to eq. [1], ΔG will be negative when |ΔH| > |T.ΔS|, that is, at low temperatures.

3 0

3 years ago

9)

Trava [24]

Answer : The volume of gas will be, 113.5 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 60 atm

$P_2$ = final pressure of gas = 30 atm

$V_1$ = initial volume of gas = 29 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = 115 K

$T_2$ = final temperature of gas = 225 K

Now put all the given values in the above equation, we get:

$\frac{60atm\times 29L}{115K}=\frac{30atm\times V_2}{225K}$

$V_2=113.5L$

Therefore, the volume of gas will be, 113.5 L

7 0

3 years ago

A confused student was doing an isomer problem and listed the following six names as different structural isomers of C7H16. a. 1

klasskru [66]

In case of heptane (C7H16) the following structural isomers are possible

shown in figure

a. 1-sec-butylpropane : this is actually 3-methyl hexane

b. 4-methylhexane : this is actually 3-methylhexane

c. 2-ethylpentane : this is actually 3-methyl hexane

d. 1-ethyl-1-methylbutane: 3-methylhexane

e. 3-methylhexane: correct IUPAC

f. 4-ethylpentane: This is actually 3-methylhexane

Hence all represent single isomer

4 0

3 years ago

Read 2 more answers

A water treatment plant has 4 settling tanks that operate in parallel (the flow gets split into 4 equal flow streams), and each

ale4655 [162]

Answer:

a) When the 4 tanks operate in parallel the retention time is 1.26 hours.

b) If the tanks are in series, the retention time would be 0.31 hours

Explanation:

The plant has 4 tanks, each tank has a volume V = 600 $m_3$ . The total flow to the plant is Ft = 12 MGD (Millions of gallons per day)

When we use the tanks in parallel, it means that the total flow will be divided in the total number of tanks. F1 will be the flow of each tank.

Firstly, we should convert the MGD to $m_3 /day$ . In that sense, we can calculate the retention time using the tank volume in $m_3$ .

$Ft =12 \frac{MG}{day} (\frac{1*10^6 gall}{1MG} ) (\frac{3.78 L}{1 gall} ) (\frac{1 dm^3}{1L} ) (\frac{1 m^3}{10^3 dm^3} ) = 45360 \frac{m^3}{day}$

After that, we should divide the total flow by four, because we have four tanks.

$F1 = Ft/4 =(45360 \frac{m^3}{d} )/4 = 11 340 \frac{m^3}{d}$

To calculate the retention time we divide the total volume V by the flow of each tank F1.

$t1 =\frac{V1}{F1} = \frac{600 m^3}{11340 \frac{m^3}{d} } = 0.0529 day\\t1 = 0.0529 d (\frac{24 h}{1d} ) = 1.26 h$

After converting t1 to hours we found that the retention time when the four reactors are in parallel is 1.26 hours.

b)

If the four reactors were working in series, the entire flow goes first through one tank, then the second and so on. It means the total flow will be the flow of each tank.

In that order of ideas, the flow for reactors in series will be F2, and will have the same value of F0.

F2 = F0

To calculate the retention time t2 we divide the total volume V by the flow of each tank F2.

$t2 =\frac{V1}{F2} = \frac{600 m^3}{45360 \frac{m^3}{d} } = 0.01 day\\t2 = 0.01 d (\frac{24 h}{1d} ) = 0.31 h$

After converting t2 to hours we found that the retention time when the four reactors are in series is 0.31 hours.

5 0

3 years ago