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Kaylis [27]
3 years ago
7

A __________ has distinct properties and composition that never vary.

Chemistry
2 answers:
WINSTONCH [101]3 years ago
7 0

Answer:

The answer is pure substance

erik [133]3 years ago
3 0
<h2>Answer </h2>

Matter

<u>Explanation </u>

A matter has distinct properties and composition that never vary. The matter is anything which possesses mass and occupies space. The matter has different characteristics and a distribution that does not change from unit to unit. Materials which cannot be disintegrated into simpler materials. Each is made of only one kind of atom in subatomic level. For example, a chair, table, and similar everything that has mass and occupies space is matter.

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4 years ago
Read 2 more answers
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
2 grams of a substance was dissolved in 75.0 g of water resulting in a temperature decrease from 23.1 ⁰C to 18.7 ⁰C. The specifi
RUDIKE [14]
<h3>Answer:</h3>

1379.4 Joules

<h3>Explanation:</h3>
  • The quantity of heat is calculated multiplying the mass of a substance by heat capacity and the change in temperature.

Therefore;

Quantity of heat = Mass × specific heat capacity × Change in temperature

Q = mcΔT

In this case;

The substance dissolved in water gained heat while water lost heat energy.

Thus, Heat gained by the substance = heat lost by water

Heat associated with the water  

Mass of water = 75 g

Change in temperature = 4.4°C

Specific heat capacity =  4.18 J/g·⁰C

Heat = mcΔT

        = 75 g × 4.18 J/g·⁰C × 4.4 °C

        =1379.4 Joules

5 0
3 years ago
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