Describe the position of metal ,non metals and metollieds in perodic tables

You need to determine the mass of an aqueous solution. You determine the mass of your 10.0 mL graduated cylinder to be 23.731 g.

Amanda [17]

Answer: 0.86g/mL

Explanation:

Mass of empty cylinder = 23.731g

Mass of cylinder + liquid = 26.414g

Mass of the liquid = 26.414 — 23.731

= 2.683g

Volume of the liquid = 3.12mL

Density = Mass / volume

Density = 2.683g / 3.12mL

Density = 0.86g/mL

3 0

3 years ago

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

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

b)

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

2 years ago

Urea, CO(NH2)2, is manufactured on a large scale for use in producing urea-formaldehyde plastics and as a fertilizer. What is th

Pachacha [2.7K]

Answer:

5.004kg

Explanation:

Combustion of carbon

C+O2=CO2

from the relationship of molar ratio

mass of carbon/molar mass of carbon=volume of CO2 produced\molar vol(22.4 dm3)

mass of carbon =1000kg

atomic mass of carbon =12

volume of CO2 produced=1000×22.4/12

volume of CO2 produced =1866.6dm3

from the combustion reaction equation provided

CO2 (g) + 2NH3 (g) ⟶ CO (NH2 )2 (s) + H2 O(l)

applying the same relationship of molar ratio

no of mole of CO2=no of mole of urea

therefore

vol of CO2\22.4=mass of urea/molar mass of urea

molar mass of urea=60.06g/mol

from the first calculation

vol of CO2=1866.6dm3

mass of urea=1866.6×60.06/22.4

mass of urea=5004.82kg

7 0

3 years ago

Read 2 more answers

The total pressure in a mixture of gases is equal to the partial pressures of

baherus [9]

Answer:

"The total pressure in a mixture of gases is equal to the sum of partial pressures of each gas"

Explanation:

Dalton's law of partial pressures state that, in a mixture of gases, the total pressure is equal to the sum of the partial pressure exerted by each gas of the mixture. The equation is:

Total pressure = Partial pressure Gas 1 + Partial pressure Gas 2 + .... + Partial pressure Gas n

To complete the sentence we can say:

"The total pressure in a mixture of gases is equal to the sum of partial pressures of each gas"

5 0

2 years ago

An atom has the following electron configuration.

Wewaii [24]

Answer: 5

Explanation: add up all the electrons and it will amount to 23. Arranging by the old model for electronic configuration, we have : 2, 8, 8, 5

The last number being 5 represent its valence electron

3 0

2 years ago

Read 2 more answers

Describe the position of metal ,non metals and metollieds in perodic tables​

Describe the position of metal ,non metals and metollieds in perodic tables