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What is the oxidation number of Nitrogen in HNO2? +1 -1 +3 -3

WITCHER [35]

I believe the correct answer from the choices listed above is the third option. The oxidation number of Nitrogen in HNO2 would be +3. It is calculated as follows:

1 + x + (-2)(2) = 0
x = +3

Hope this answers the question. Have a nice day.

5 0

3 years ago

Which of the following pairs are about equal in mass?

Mademuasel [1]

Answer:

Proton and neutrons

Explanation:

they both have a mass of 1 amu or 1 atomic mass

3 0

3 years ago

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How many particles are present in 12.47 grams of NaCl

liq [111]

Answer:

1.26*10²³ particles are present in 12.47 grams of NaCl

Explanation:

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023 * 10²³ particles per mole. The Avogadro number applies to any substance.

So, first of all you must know the amount of moles that represent 12.47 grams of NaCl. For that it is necessary to know the molar mass.

You know:

Na: 23 g/mole
Cl: 35.45 g/mole

So the molar mass of NaCl is: 23 g/mole + 35.45 g/mole= 58.45 g/mole

Now you apply a rule of three as follows: if 58.45 grams are present in 1 mole of NaCl, 12.47 grams in how many moles will they be?

$moles=\frac{12.47 grams*1 mole}{58.45 grams}$

moles= 0.21

You apply a rule of three again, knowing Avogadro's number: if in 1 mole of NaCl there are 6,023 * 10²³ particles, in 0.21 moles how many particles are there?

$number of particles=\frac{0.21 moles*6.023*10^{23} }{1 mole}$

number of particles= 1.26*10²³

1.26*10²³ particles are present in 12.47 grams of NaCl

5 0

3 years ago

Ground water, the largest source of fresh water, is stored in bodies of rock and or sediment

zepelin [54]

D. Aquifers
Hope this helps chief.

7 0

3 years ago

A single-effect evaporator is concentrating a feed of 9072 kg/h of a 10 wt % solution of NaOH in water at temperature of 288.8 K

Aloiza [94]

Answer:

a) steam used = 8440 kg/hr

b) Because as per calculation done in a), Steam is sufficient to evaporate 7257.6 kg/hr water from feed. Since feed is less than this amount, therefore whole water will be evaporated and 100% solid product will be obtained in bottom collection

Explanation:

Saturated steam pressure = 42KPag=1.42 bar

From steam table, Steam temperature = 110 C

Latent heat of this steam = 2230 KJ/kg

Process side pressure = 20 Kpaa = .2 bar

Water latent heat at this Pressue = 2360 KJ/kg

Water boiling Point at this Pressure =60 C

Feed Inlet temperature = 15.6 C

Total Heat required = Heat required to rise feed temperature from 15.6 to 60 degree C + Evaporation of water to concentrate the feed

Total Water evaporation required = 9072*(100-10)/100-9072*.1/.5*.5=7257.6 Kg/hr

Specific heat of feed assumed = 4.2 KJ/kg/K

=> Total heat required = 9072*4.2*(60-15.6)+7257.6*2360=18819682 KJ/hr

LMTD = ((110-60)-(110-15.6))/LN((110-60)/(110-15.6))=69.8 C

U, Overall Heat transfer coefficient = 1988 W/m2/K

Total heat required = U*A*LMTD

=> Area required of evaporator, A=18819682 *1000/3600/1988/59.8=44 m2

Steam used = 18819682/2230=8440 kg/hr

Energy required to condense vaporised feed = 7257.6*2360=17127936 KJ/hr

=> Steam efficiency = (18819682-17127936)/18819682=9%

b) Because as per calculation done in a), Steam is sufficient to evaporate 7257.6 kg/hr water from feed. Since feed is less than this amount, therefore whole water will be evaporated and 100% solid product will be obtained in bottom collection

6 0

3 years ago

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