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Elden [556K]
3 years ago
11

How many moles of solute

Chemistry
1 answer:
nydimaria [60]3 years ago
6 0

Answer:

3.4483 ok I googled it

Explanation:

h hhejeje

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Which of the following results from acid deposition? View Available Hint(s) Which of the following results from acid deposition?
ASHA 777 [7]

Answer:

Objects made of stone, like marble headstones, are worn away.

Explanation:

Acid deposition on earth is mostly as a result of acid rain. Acid rain forms which oxides of non-metals such as nitrogen and carbon dissolves in rain water to produce weak acid.

This weak acid that is produced easily wears away materials made up of marble and other sculptures. The most active of the acid is weak carbonic acid produced by dissolution of carbon dioxide in rain water.

6 0
2 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
Read 2 more answers
1- A sample of gas is in a 3.00 L contain at a pressure of 740.0 mmHg. What is the new pressure of the sample if the container's
inna [77]

Considering the Boyle's law, the new pressure of the sample is 1,776 mmHg.

<h3>What is Boyle's law</h3>

Boyle's law establishes the relationship between the pressure and the volume of a gas when the temperature is constant.

Boyle's law states that the volume occupied by a given mass of gas at constant temperature is inversely proportional to the pressure. This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

P×V=k

Now it is possible to assume that you have a certain volume of gas V1 which is at a pressure P1 at the beginning of the experiment. If you vary the volume of gas to a new value V2, then the pressure will change to P2, and the following will be true:

P1×V1=P2×V2

<h3>New pressure</h3>

In this case, you know:

  • P1= 740 mmHg
  • V1= 3 L
  • P2= ?
  • V2= 1.25 L

Replacing in Boyle's law:

740 mmHg× 3 L=P2× 1.25 L

Solving:

P2= (740 mmHg× 3 L) ÷ 1.25 L

P2= 1,776 mmHg

Finally, the new pressure of the sample is 1,776 mmHg.

Learn more about Boyle's law:

brainly.com/question/4147359

#SPJ1

7 0
2 years ago
Read 2 more answers
The student recorded the mass of the cup + sample incorrectly and started with 2.20 g of hydrated compound but used 2.00 g in th
gtnhenbr [62]

Answer:

low

Explanation:

We were informed in the question that the student had incorrectly recorded the mass of cup + sample as 2.20 g but inadvertently used 2.00 g in the calculations.

This error will cause a slight decrease in the mass of water and ultimately decrease the number of moles of water in the hydrate.

What i am saying is that the number of moles of water obtained in the calculation will be artificially low.

4 0
2 years ago
The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How
BARSIC [14]
<h3>Answer:</h3>

              6.21 × 10²² Carbon Atoms

<h3>Solution:</h3>

Data Given:

                 Mass of Butane (C₄H₁₀)  =  1.50 g

                 M.Mass of Butane  =  58.1 g.mol⁻¹

Step 1: Calculate Moles of Butane as,

                 Moles  =  Mass ÷ M.Mass

Putting values,

                 Moles  =  1.50 g ÷ 58.1 g.mol⁻¹

                 Moles  =  0.0258 mol

Step 2: Calculate number of Butane Molecules;

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of Butane Molecules can be written as,

            Moles  =  Number of C₄H₁₀ Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of Butane molecules,

             Number of C₄H₁₀ Molecules  =  Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting value of moles,

     Number of C₄H₁₀ Molecules  =  0.0258 mol × 6.022 × 10²³ Molecules.mol⁻¹

                 Number of C₄H₁₀ Molecules  =  1.55 × 10²² C₄H₁₀ Molecules

Step 3: Calculate Number of Carbon Atoms:

As,

                            1 Molecule of C₄H₁₀ contains  =  4 Atoms of Carbon

So,

          1.55 × 10²² C₄H₁₀ Molecules will contain  =  X Atoms of Carbon

Solving for X,

 X =  (1.55 × 10²² C₄H₁₀ Molecules × 4 Atoms of Carbon) ÷ 1 Molecule of C₄H₁₀

X  =  6.21 × 10²² Atoms of Carbon

5 0
3 years ago
Read 2 more answers
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