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Dennis_Churaev [7]
2 years ago
7

Consider pure water separated from an aqueous sugar solution by a semipermeable membrane, which allows water to pass freely but

not sugar. After some time has passed, the concentration of sugar solution: a. will have decreased b. will be the same on both sides of the membrane c. might have increased or decreased depending on other factors d. will not have changed e. will have increased
Chemistry
1 answer:
mojhsa [17]2 years ago
7 0

Answer:

it will be the same on both sides of the membrane

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The energy released by a nuclear fusion reaction is produced when
Igoryamba

Answer:

2

Explanation:

energy produced in nuclear fusion or even fission is due to mass defect and it can be calculated by Einstein equation e=mc2

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3 years ago
Balance <br><br> _Mg + _HCL = _MgCl2 + H2
Pavel [41]
_Mg + _HCL = _MgCl2 + H2
Separate the terms on each side: 

_Mg + _HCl = _MgCl2 + H2

Mg- 1                 Mg-1
               
H-1                    H-2

Cl-1                    Cl-2

Mg is balanced on both sides so move on to the next (put a 1 in the space). 

1Mg

There are two H's and two Cl's on the results side, so to balance the equation put a 2 as a coefficient for HCl and it'll all balance out.

2HCl

Balamced equation will be: 

 1Mg + 2HCL = 1MgCl2 + H2
5 0
3 years ago
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Which of the following is an example of point-source pollution? A. Salt from roads B. Water from a sewage treatment plant C. Fer
ivanzaharov [21]

Answer:

Acid from abandoned mines- D.

7 0
3 years ago
The molecule shown belongs to which class of compounds
Kaylis [27]
The answer is (4) amino acid. This molecule has one carboxyl and one amidogen linked at the same carbon atom. This is the property of amino acid. So this is an amino acid.
7 0
3 years ago
1. The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3⋅xH2O The number of kilowatt-hour
lianna [129]

*Answer:

Option A: 59.6

Explanation:

Step 1: Data given

Mass of aluminium = 4.00 kg

The applied emf = 5.00 V

watts = volts * amperes

Step 2: Calculate amperes

equivalent mass of aluminum = 27 / 3 = 9  

mass of deposit = (equivalent mass x amperes x seconds) / 96500

4000 grams = (9* amperes * seconds) / 96500

amperes * seconds = 42888888.9

1 hour = 3600 seconds

amperes * hours = 42888888.9 / 3600 = 11913.6

amperes = 11913.6 / hours

Step 3: Calculate kilowatts

watts = 5 * 11913.6 / hours

watts = 59568 (per hour)

kilowatts = 59.6 (per hour)

The number of kilowatt-hours of electricity required to produce 4.00kg of aluminum from electrolysis of compounds from bauxite is 59.6 kWh when the applied emf is 5.00V

5 0
3 years ago
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