Why does a chemical equation be balanced?
1 answer:
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Answer:
Part A:
Charge is
Configuration is
Part B:
Charge is
Configuration is
Part C:
Charge is
Configuration is
Explanation:
Monatomic ions:
These ions consist of only one atom. If they have more than one atom then they are poly atomic ions.
Examples of Mono Atomic ions:
Part A:
For P:
Phosphorous (P) has 15 electrons so it require 3 more electrons to stabilize itself.
Charge is
Full ground-state electron configuration of the mono atomic ion:
Part B:
For Mg:
Magnesium (Mg) has 12 electrons so it requires 2 electrons to lose to achieve stable configuration.
Charge is
Full ground-state electron configuration of the mono atomic ion:
Part C:
For Se:
Selenium (Se) has 34 electrons and requires two electrons to be stable.
Charge is
Full ground-state electron configuration of the mono atomic ion:
Answer:
Explanation:
Hello!
In this case, since the percent water is computed by dividing the amount of water by the total mass of the hydrate; we infer we first need the molar mass of water and that of the hydrate as shown below:
Thus, the percent water is:
So we plug in to obtain:
Best regards!
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 :-)
