answer:
the student <u>who</u> answers the riddle will get the prize
explanation:
- who is the pronoun
- a pronoun is something that substitutes for a noun
Answer:
Explanation:
Relation between ΔG₀ and K ( equilibrium constant ) is as follows .
lnK = - ΔG₀ / RT
The value of R and T are same for all reactions .
So higher the value of negative ΔG₀ , higher will be the value of K .
Mg(s) + N₂0(g) → MgO(s) + N₂(g)
has the ΔG₀ value of -673 kJ which is highest negative value . So this reaction will have highest value of equilibrium constant K .
Number of coulombs of positive charge in 250cm^3 water is 1.3×10^7 C
The volume of 250 cm^3 corresponds to a mass of 250 g since the density of water is 1.0 g/cm^3
This mass corresponds to 250/18 = 14 moles since the molar mass of water is 18. There are ten proton (each with charge q = +e) in each molecule of 
So,
Q = 14NA q =14(6.02×10^23)(10)(1.60×10^−19C) = 1.3×10^7 C.
Mass is the quantity of matter in a physical body. It is also a measure of the body's inertia, the resistance to acceleration when a net force is applied. An object's mass also determines the strength of its gravitational attraction to other bodies.
Learn more about mass here:
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The volume of SO2 produced at 325k is calculated as below
calculate the moles of SO2 produced which is calculated as follows
write the reacting equation
K2SO3 +2 HCl = 2KCl +H2O+ SO2
find the moles of HCl used
=mass/molar mass = 15g/ 36.5 g/mol =0.411 moles
by use of mole ratio between HCl to SO2 which is 2:1 the moles of SO2 is therefore = 0.411 /2 =0.206 moles of SO2
use the idea gas equation to calculate the volume SO2
that is V=nRT/P
where n=0.206 moles
R(gas constant) = 0.082 L.atm/ mol.k
T=325 K
P=1.35 atm
V=(0.206 moles x 0.082 L.atm/mol.k x325 k)/1.35 atm = 4.07 L of SO2
The balanced combustion reaction of propane, C₃H₈, is
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
Molar mass of propane: 44 g/mol
Moles of propane = 42 g * (1 mol/44g) = 0.9545 mol propane
Molar mass of oxygen: 32 g/mol
Moles of oxygen = 115 g * (1 mol/32 g) = 3.594 mol oxygen
Moles of oxygen needed to completely react propane:
0.9545 mol propane * (5 mol O₂/1 mol propane) = 4.7725 mol oxygen
Since the available oxygen is only 3.594 moles and propane needs 4.7725 moles, that means oxygen is our limiting reactant. We base the amount of water produced here.
Molar mass of water: 18 g/mol
Mass of water produced = 3.594 mol O₂ * (4 mol H₂O/5 mol O₂) * (18 g/mol)
Mass of water produced = 258.768 grams
