Complete the equation for respiration: Oxygen + ________ ---> carbon dioxide + energy + water *
1 answer:
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Answer:
Na₂CO₃.2H₂O
Explanation:
For the hydrated compound, let us denote is by Na₂CO₃.xH₂O
The unknown is the value of x which is the amount of water of crystallisation.
Given values:
Starting mass of hydrate i.e Na₂CO₃.xH₂O = 4.31g
Mass after heating (Na₂CO₃) = 3.22g
Mass of the water of crystallisation = (4.31-3.22)g = 1.09g
To determine the integer x, we find the number of moles of the anhydrous Na₂CO₃ and that of the water of crystallisation:
Number of moles =
Molar mass of Na₂CO₃ =[(23x2) + 12 + (16x3)] = 106gmol⁻¹
Molar mass of H₂O = [(1x2) + (16)] = 18gmol⁻¹
Number of moles of Na₂CO₃ = = 0.03mole
Number of moles of H₂O = = 0.06mole
From the obtained number of moles:
Na₂CO₃ H₂O
0.03 0.06
Simplest
Ratio 0.03/0.03 0.03/0.06
1 2
Therefore, x = 2
Answer:
B
Explanation:
hope the picture helps you to understand:)
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 :-)
