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grin007 [14]
3 years ago
6

The amount of heat needed to raise 25 g of a substance by 15°C is 293 J. What is the specific heat of the substance? Use the equ

ation C = q/mΔT.
a.) 0.13 J/g-°C

b.) 0.20 J/g-°C

c.) 0.78 J/g-°C

d.) 0.46 J/g-°C
Chemistry
1 answer:
balandron [24]3 years ago
3 0
It is so simple and direct substitution in the formula of :
C= Q/ m ΔT
when C is specific heat constant
and Q is heat energy of joul 
and m is the mass
and ΔT is the change in temperature
and when we have Q = 293 J & m = 25 g & ΔT = 15 °C so by substitution:
∴C = 293/(25*15)
      = 0.78 J/g-°C 
∴ 0.78 J/g-°C is the correct answer

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Naddik [55]
The answer is acidic 
6 0
3 years ago
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hichkok12 [17]

Answer:

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6 0
3 years ago
How many grams of NaCl (molecular weight = 58 Da) would you have to dissolve in 50ml to make a 2M stock solution? Answer:
Lerok [7]

Answer:

5.8 g

Explanation:

Molecular weight in Daltons is equivalent to the molecular weight in grams per mole.

The amount of NaCl required is calculated as follows:

(2 mol/L)(50 mL)(1 L/1000 mL) = 0.1 mol

This amount is converted to grams using the molar mass (58 g/mol).

(0.1 mol)(58 g/mol) = 5.8 g

7 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 mass of sodium chloride (NaCl) forms when 7.5 g of sodium carbonate (Na2CO3) reacts with a dilute solution of hydrochloric
Gre4nikov [31]

 The mass of NaCl  formed  is 8.307  grams


<u><em> calculation</em></u>

step 1: write the equation  for reaction

Na₂CO₃  + 2HCl → 2 NaCl  +CO₂ +H₂O

Step 2: find the  moles of Na₂CO₃

moles = mass/molar mass

 The  molar mass of Na₂CO₃  is = (23 x2) + 12 + ( 16 x3) = 106 g/mol

moles  = 7.5 g/106 g/mol =0.071 moles

Step 3: use the  mole ratio to determine the  mole of NaCl

Na₂CO₃:NaCl  is  1:2  therefore the moles of NaCl =0.07  x2 =0.142 moles


Step 4:  calculate mass  of NaCl

mass= moles x molar mass

the molar  mass of NaCl= 23 +35.5 =58.5 g/mol

mass  = 0.142  moles x 58.5 g/mol =8.307  grams

8 0
3 years ago
Read 2 more answers
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