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1 year ago

Write balanced equations and Kb expressions for these Brønsted-Lowry bases in water:(CH₃)₃N

Chemistry

1 answer:

grandymaker [24]1 year ago

8 0

(CH3)3N + H2O ---> (CH3)3NH+ + OH-

The equation of Kb:

Kb = [(CH3)3NH+][OH-]/[(CH3)3N]

<h3>What do we know about the process of dissolution?</h3>

A solute can dissolve in a solvent in a solid, liquid, or gaseous phase to create a solution through the process of dissolution. Solubility. The greatest amount of a solute that may dissolve in a solvent at a particular temperature is known as its solubility. The solution is said to as saturated when the maximum amount of solute is present.

Several variables influence solubility, including:

The solute's concentration
The system's temperature
Pressure (for gases in solution)
The solvent's and solute's polarity

To learn more about pKb:

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A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt

Nadusha1986 [10]

Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

$dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt$

integrating factor

$=e^{\int\limits (1/40) \, dt}$

integrating factor $=e^{(1/40)t}$

multiply on both sides by $=e^{(1/40)t}$

$dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t$

integrate on both sides

$\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})$

after long period of time means t - > ∞

${t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80$

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

6 0

3 years ago

Any element above 92 on the

Fynjy0 [20]

False
explanation

All of the elements with atomic numbers 1 to 92 can be found in nature, have stable or very long half-life isotopes, and are created as common products of the decay of uranium and thorium.

5 0

2 years ago

Read 2 more answers

Tim wants to store 0.122 moles of helium gas at 203 kPa and 401 K. Which is the volume of the helium gas needed for storage?

algol13

The volume of Helium gas needed for storage is 2.00 L (answer C)

calculation

The volume of Helium is calculated using ideal gas equation

That is Pv =nRT

where;

P( pressure) = 203 KPa

V(volume)=?

n(number of moles) = 0.122 moles

R(gas constant) = 8.314 L.Kpa/mol.K

T(temperature)= 401 K

make V the subject of the formula by diving both side by P

V=nRT/p

V={[0.122 moles x 8.314 L. KPa/mol.K x 401 K] / 203 KPa} = 2.00 L

3 0

3 years ago

A mixture of nacl and sucrose (c12h22o11) of combined mass 10.2 g is dissolved in enough water to make up a 250 ml solution. the

Vilka [71]

We use the osmotic pressure to determine the concentration of the solute in the solution. Then, we multiply the volume of the solution to determine the number of moles of solute particles. We need to establish to equations since we have two unknowns, the mass of of each solute. We do as follows:

osmotic pressure = CRT
C = 7.75 / 0.08205 (296.15) = 0.3189 mol / L
moles of particles = C*V = 0.3189*0.250 =0.0797 mol 
0.0797 = moles of sucrose + 2*moles of salt 

x + 2y = 0.0797 
and 
x(MMsucrose) + y(MMNaCl) = 10.2
342x + 58.5y = 10.2

solve for x and y

x = 0.0252 mol sucrose
y = 0.0273 mol NaCl

mass Sucrose = 0.0252(342) = 8.6184 g 
mass NaCl = 0.0273(58.5) = 1.5971 g 
% NaCl = (1.5971 / 10.2)*100 = 15.66%

4 0

3 years ago

A tow truck does 10,000 J of work in 5 seconds. Calculate the power.

chubhunter [2.5K]

Answer:

power is work/time so 10,000j/5s=2000J/s

5 0

3 years ago