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OLga [1]
3 years ago
7

How many joules of heat are released when 33 g of water cools from 77°C to

Chemistry
1 answer:
lubasha [3.4K]3 years ago
5 0

Answer:

10207.56 j

Explanation:

have a nice day

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Ksenya-84 [330]

The disease can slow the predator down and the prey can get away easily.

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What was the key characteristic by which Dmitri Mendeleev arranged his periodic table? A. increasing atomic number B. electron c
valentina_108 [34]
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4 0
3 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
The buffer is prepared by adding 250mL of 0.80M NH3 to 150mL of 0.50M NH4NO3. What is the pH of the final solution (Kb for NH3 =
madam [21]

The pH of the solution is obtained to be 9.67.

<h3>What is a buffer?</h3>

A buffer is a solution that serves to militate against changes in acidity or alkalinity. The buffer is often made of a solution of a strong acid and its salt or a weak base and its salt.

Now we have the that;

Number of moles of base = 250 /1000 *  0.80 = 0.2 moles

Number of moles of salt = 150/1000 *  0.50 = 0.075 moles

Total volume of the solution = 250 + 150 = 400 mL or 0.4 L

Molarity of the base = 0.2 moles/0.4 L = 0.5 M

Molarity of the salt =  0.075 moles/ 0.4 L = 0.1875 M

pKb = - log( 1.8 x 10^-5) = 4.74

pOH = pKb + log [salt/base]

pOH =  4.74 + log (0.1875/ 0.5)

pOH = 4.33

pH = 14 - 4.33

pH = 9.67

The pH of the solution is obtained to be 9.67.

Learn more about pH:brainly.com/question/15289741

#SPJ1

6 0
2 years ago
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