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BartSMP [9]
3 years ago
7

Which of the responses contains all the true

Chemistry
1 answer:
jekas [21]3 years ago
7 0
What responses am i supposed to chose from?
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Which of the following is not true of alloys?
Scilla [17]
Option d: copper.

Because copper is an element, not a mixture.

7 0
4 years ago
Here anwser questions 2 ,4,6 those charts okay I also put how the formula on top solve these problems ! Please help me I geniunu
disa [49]

Answer:

39.5L

Explanation:

4 0
3 years ago
WHAT WILL HAPPEN TO THE RESISTANCE OF CONDUCTOR IF :
Debora [2.8K]

Answer:

1. The length of the wire is halved.

the resistance of a conductor is directly proportional to it's length (L) as R - L. Thus doubling it's length will double it's resistance, while halving it's length would halve the resistance. Also the resistance of a conductor is inversely proportional to it's cross sectional area.

2. The area of cross-section of the conductor in increased.

On increasing the area of cross-section, resistance decreases. Thus is because resistance is inversely proportional to area.

3. The temperature of the conductor is increased.

With increasing temperature, the resistance of the wire increases as collisions within the wire increases and "slow" the flow of current... Since conductors typically display an increased resistivity with temperature increase, they have a positive temperature coefficient.

3 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
What kind of chemical reaction these equations represent.
Sedbober [7]
I believe it would be combustion because of the products CO2 and then C+2H2
7 0
3 years ago
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