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zzz [600]
3 years ago
10

When floral design first emerged as an art form, which technique was emphasized?

Chemistry
1 answer:
olasank [31]3 years ago
8 0
Using flowers in their natural state and shape
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A bottle of 2 moles of an ideal diatomic gas experiences a temperature increase of 50.0K at constant volume. (a) Find the increa
andre [41]

Answer:

Explanation:

a) For diatomic gas: Translational motion = 3 and rotational motion = 2

∴ Total (internal energy) = 3 + 2 = 5

b) Translational + Rotational + Vibrational = 3 + 2 + 1 = 6

c) Linear molecule  

i) Non linear molecule

ii)  Monatomic molecule

5 0
3 years ago
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
True or False: As the ice warms up, the molecules get more energy and move around more from place to place to form liquid water.
Arada [10]
The answer is TRUE.

Hope it helps.
7 0
3 years ago
Read 2 more answers
Dribbling a basketball harder into the floor makes it bounce higher is an example of what newton's law?
vodomira [7]
Newton’s third law, because a person(a) is acting upon the ball(b) by dribbling the ball on the floor
8 0
3 years ago
Why is melting point considered a physical property
miskamm [114]

Answer:

Because you can physically see the object melting when it comes to the melting point. The objects texture, color, temperature, shape, and state of matter (solid, liquid, gas) are possibly changing.

6 0
3 years ago
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