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Sindrei [870]
3 years ago
5

Solution 1 contains 3.5g of sodium chloride in 500 mL of water. Solution 2 contains 7.5g of glucose in 300 mL of water. If the a

bove solutions were separated by a membrane permeable to water, but not to the solutes, would osmosis occur, and if so, in which direction would water move?
Chemistry
1 answer:
meriva3 years ago
8 0

Answer:

Answer is Object 2 (which has a density of 1.9 g/cm³).

Explanation;

   When object is floating, the weight of that object is less than the up thrust on it.

   When an object fully submerged and floating, then the weight of that object is equal to the up thrust on it.

   This is known as the Archemide's principle.

   Both up thrust and weight depends on the density. Hence, if the density of the solution is high, then the up thrust also high. If the density high, the the weight of the object also high.

   Hence, to sink the object in water, that object should be denser than water. Hence, answer is object 3 which has a higher density than water.

Explanation:

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Any help would be appreciated. Confused.
masya89 [10]

Answer:

q(problem 1) = 25,050 joules;  q(problem 2) = 4.52 x 10⁶ joules

Explanation:

To understand these type problems one needs to go through a simple set of calculations relating to the 'HEATING CURVE OF WATER'. That is, consider the following problem ...

=> Calculate the total amount of heat needed to convert 10g ice at -10°C to steam at 110°C. Given are the following constants:

Heat of fusion (ΔHₓ) = 80 cal/gram

Heat of vaporization (ΔHv) = 540 cal/gram

specific heat of ice [c(i)] = 0.50 cal/gram·°C

specific heat of water [c(w)] = 1.00 cal/gram·°C

specific heat of steam [c(s)] = 0.48 cal/gram·°C

Now, the problem calculates the heat flow in each of five (5) phase transition regions based on the heating curve of water (see attached graph below this post) ...   Note two types of regions (1) regions of increasing slopes use q = mcΔT and (2) regions of zero slopes use q = m·ΔH.

q(warming ice) =  m·c(i)·ΔT = (10g)(0.50 cal/g°C)(10°C) = 50 cal

q(melting) = m·ΔHₓ = (10g)(80cal/g) 800 cal

q(warming water) = m·c(w)·ΔT = (10g)(1.00 cal/g°C)(100°C) = 1000 cal

q(evaporation of water) =  m·ΔHv = (10g)(540cal/g) = 5400 cal

q(heating steam) = m·c(s)·ΔT = (10g)(0.48 cal/g°C)(10°C) = 48 cal

Q(total) = ∑q = (50 + 800 + 1000 + 5400 + 48) = 7298 cals. => to convert to joules, multiply by 4.184 j/cal => q = 7298 cals x 4.184 j/cal = 30,534 joules = 30.5 Kj.

Now, for the problems in your post ... they represent fragments of the above problem. All you need to do is decide if the problem contains a temperature change (use q = m·c·ΔT) or does NOT contain a temperature change (use q = m·ΔH).    

Problem 1: Given Heat of Fusion of Water = 334 j/g, determine heat needed to melt 75g ice.

Since this is a phase transition (melting), NO temperature change occurs; use q = m·ΔHₓ = (75g)(334 j/g) = 25,050 joules.

Problem 2: Given Heat of Vaporization = 2260 j/g; determine the amount of heat needed to boil to vapor 2 Liters water ( = 2000 grams water ).

Since this is a phase transition (boiling = evaporation), NO temperature change occurs; use q = m·ΔHf = (2000g)(2260 j/g) = 4,520,000 joules = 4.52 x 10⁶ joules.

Problems containing a temperature change:

NOTE: A specific temperature change will be evident in the context of problems containing temperature change => use q = m·c·ΔT. Such is associated with the increasing slope regions of the heating curve.  Good luck on your efforts. Doc :-)

5 0
3 years ago
What is the difference between physical and chemical change?
Vikki [24]
There are several differences between<span> a </span>physical and chemical change<span> in matter or substances. A </span>physical change<span> in a substance doesn't </span>change<span> what the substance is. In a </span>chemical change<span> where there is a </span>chemical<span> reaction, a new substance is formed and energy is either given off or absorbed.</span>
6 0
3 years ago
Read 2 more answers
A sample of nitrogen gas, (N2), occupies 45.0 mL at 27.00∘C and 80.0 kPa. What will be the pressure if the gas is cooled to −73.
maxonik [38]

Answer:

The new pressure is 53.3 kPa

Explanation:

This problem can be solved by this law. when the volume remains constant, pressure changes directly proportional as the Aboslute T° is modified.

T° increase → Pressure increase

T° decrease → Pressure decrease

In this case, temperature was really decreased. So the pressure must be lower.

P₁ / T₁ = P₂ / T₂

80 kPa / 300K =  P₂/200K

(80 kPa / 300K) . 200 K = P₂ →  53.3 kPa

6 0
3 years ago
How many bonded pairs of electrons are there in the molecule H2CO?
Drupady [299]
There are eight bonded electrons in this molecule! :)
8 0
3 years ago
Which components of the apparatus affect its motion?
vodka [1.7K]

Answer:

An internal force acts between elements of the system. ... Only external forces affect the motion of a system, according to Newton's first ... Newton's second law states that a net force on an object is responsible for its ... Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological

Explanation:

5 0
3 years ago
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