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Usimov [2.4K]
2 years ago
13

In lab, a dialysis bag was filled with a 10% sucrose solution and placed in an unlabeled beaker filled with clear liquid. The di

alysis bag was made of a semipermeable membrane that allowed the free passage of water, but was not permeable to sucrose. After two hours, the bag in the beaker decreased in size, lost volume, and became flaccid. Therefore, at the beginning of the experiment, the solution in the bag was ____________ compared to the solution in the beaker. g
Chemistry
1 answer:
DaniilM [7]2 years ago
5 0

Answer:

Of lower concentration or less concentrated

Explanation:

Osmosis is the movement of solvent from a region of lower concentration of solute to a region of higher concentration of solute through a semipermeable membrane in order to equalize the concentration of the solutions on both sides.

Since the membrane of the bag is semipermeable, then the fact that the bag in the beaker decreased in size, lost volume, and became flaccid indicates that the solution in the bag is of lower solute concentration than the solution in the beaker hence the movement of water molecules into the beaker by osmosis.

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In which two types of reactions do opposite chemical changes occur
RideAnS [48]

Hello!

Before I can answer your question we need to know the different types of chemical reactions that can occur:

  • Single Replacement
  • Double Replacement
  • Synthesis
  • Decomposition
  • Combustion

Combustion is an answer that can usually be ruled out unless oxidation is occuring in the chemical equation (hyrdrogen occuring with oxygen to react)

Synthesis reaction, also known as a Combination reaction, is when the reactants combine to form a single product.

Decompostion is when a single reactant is deconstructed to form multiple products, or the opposite of Synthesis.


With these definitions alone, since they are opposite types of reactions, the answer would be Synthesis and Decomposition reactions !

Hope this helps and if you need anymore clarification feel free to ask.


4 0
3 years ago
Compare the keys to periodic tables a and b to the key to the periodic table in your textbook. what other information is provide
lesya692 [45]
What the definition of each period and group and element and every valence electron
5 0
3 years ago
What is the total pressure in a container with 256 mm Hg of Oz, 198 mm Hg of He,
irina1246 [14]

Answer:

P(total pressure) = 504 mmHg = 504mm/760mm/atm = 0.663 atm

Explanation:

Apply Dalton's Law of Partial Pressures.

P(total) = ∑Partial Pressures = ∑(256mm + 198mm + 48mm) = 504 mmHg

P(total pressure) = 504 mmHg = 504mm/760mm/atm = 0.663 atm

7 0
2 years ago
If 5.57 g of Ag2O is sealed in a 75.0-mL tube filled with 760 torr of N2 gas at 28 ∘C, and the tube is heated to 310 ∘C, the Ag2
Snezhnost [94]

Answer: Total pressure = 7293.2 torr or 9.60 atm

Explanation:

<em>Total pressure = partial pressure of nitrogen + partial pressure of oxygen</em>

The partial pressure due to nitrogen is determined using the equation of Gay-Lussac's law: <em>P₁/T₁=P₂/T₂</em>

P₁ = 760 torr = 1atm, T₁ = 28∘C = (273+28)K = 301k, P₂ = ?, T₂ = 310∘C =(310+273)K = 583K

P₂ = P₁ T₂/ T₁

P₂ = 760 * 583 / 301 = 1472.03 torr

The pressure due to Oxygen gas produced is calculated thus:

Balanced equation of the decomposition of Ag₂O at s.t.p. is as follows;

2Ag₂O ----> 4Ag + O₂(g)

2 moles of Ag₂O produces 1 mole of O₂

molar mass of Ag₂O = (2*108 + 16)g = 232g/mol

molar volume of gas at s.t.p. = 22.4L

2*232g i.e. 464g of Ag₂O produces 22.4L of O₂

5.57g of Ag₂O will produce 5.57g*22.4L/464g = 0.269L or 269mL of O₂

Using the General gas equation  P₁V₁/T₁=P₂V₂/T₂

P₁ = 1atm = 760 torr, V₁ = 269mL, T₁=273K, P₂ = ?, V₂= 75mL, T₂ = 583K

P₂ = P₁V₁T₂/V₂T₁

P₂ = 760*269*583 / 75*273

P₂ = 5821.17 torr

Total pressure = (1472.03 + 5821.17) torr

Total pressure = 7293.2 torr or 9.60 atm

7 0
2 years ago
Calculate the volume of 38.0 g of carbon dioxide at STP. Enter your answer in the box provided. L
Free_Kalibri [48]

Answer:

19.3 L

Explanation:

V= n × 22.4

where V is volume and n is moles

First, to find the moles of CO2, divide 38.0 by the molecular weight of CO2 which is 44.01

n= m/ MM

n= 38/ 44.01

n= 0.86344012724

V= 0.86344012724 × 22.4

V= 19.3410588502 L

V= 19.3 L

7 0
3 years ago
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