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Svet_ta [14]
3 years ago
15

Given the balanced equation representing a reaction:

Chemistry
1 answer:
Hunter-Best [27]3 years ago
8 0
The answer is (4) a substitution reaction. Notice that propane lost an H and gained a Br. This constitutes substitution.
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PLEASE HELP!!!!!!!
Basile [38]

Answer:

2nd option

Explanation:

5 0
3 years ago
An element's atomic number is 86. how many protons would an atom of this element have?
rusak2 [61]
The number of protons an element has is the same as the atomic number, so the element would have 86 protons.
8 0
3 years ago
Give the orbital notation for a neutral atom of titanium
dedylja [7]

Answer:

Explanation:

A neutral titanium atom will have 22 electrons. Therefore, its electron configuration will be 1s2 2s2 2p6 3s2 3p6 4s2 3d2.

5 0
2 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
If your car gets 30.3 mi/gal, how many gallons of gasoline would you use if you drove 506.3 miles? 1 mile = 5280 ft (exactly) In
Salsk061 [2.6K]

Answer: 16.7 gallon

Explanation:

Given: The car can drive 30.3 miles when 1 gallon of gasoline is used.

Distance covered = 506.3 miles

Thus for 30.3 miles, the amount of gasoline used= 1 gallon

For 506.3 miles, the amount of gasoline used=\frac{1}{30.3}\times 506.3=16.7gallon

Thus the amount of gasoline used is 16.7 gallons.


4 0
3 years ago
Read 2 more answers
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