Sr will always have 38 protons because that is its atomic number. Normally Sr has 38 protons and 49 neutrons yielding an atomic mass of 87. unless you are dealing with an isotope of Strontium in which 3 additional neutrons would give you the 53 neutrons in your answer, (38p, 52n).
Answer:
Sodium oxalate is a basic salt. In water it can be dissolved and dissociated.
The oxalic acid in water has two dissociations.
Explanation:
Na2C2O4 ---> 2Na+ + C2O4-2
Sodium oxalate is the conjugate base of a weak acid. In water this salt, dissociates completely giving rise to the sodium and oxalate ions. As Na+ comes from a strong base, in water it does not produce hydrolysis while oxalate does react in water, because it takes a proton from it and it generates a basic hydrolysis releasing OH-.
C2O4-2 + H2O ⇄ HC2O4- + OH-
In water the salt is basic. The pH of an aqueous solution of this salt is basic, since OH- is generated.
The HC2O4- has a second hydrolisis, it takes another proton from water to form oxalic acid.
HC2O4- + H2O ⇄ H2C2O4 + OH-
The oxalic acid acts as a weak acid, it can release 2 protons to water, to make oxalate (its conjugate base).
H2C2O4 + H2O ⇄ H3O+ + HC2O4-
HC2O4- + H2O ⇄ H3O+ C2O4-2
The HC2O4- acts as an ampholyte since it accepts and delivers protons simultaneously.
Answer:
x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates
Explanation:
Recall that , depression present in freezing point is calculated with the formulae = solute particles Molarity x KF
0.3473 = m * 1.86
Solving, m = 0.187 m
Moles of HClO2 = mass / molar mass = 5.85 / 68.5 = 0.0854 mol
Molality = moles / mass of water in kg = 0.0854 / 1 = 0.0854 m
Initial molality
Assuming that a % x of the solute dissociates, we have the ICE table:
HClO2 H+ + ClO2-
initial concentration: 0.0854 0 0
final concentration: 0.0854(1-x/100) 0.0854x/100 0.0854x / 100
We see that sum of molality of equilibrium mixture = freezing point molality
0.0854( 1 - x/100 + x/100 + x/100) = 0.187
2.1897 = 1 + x / 100
x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates
Each enzyme's active site is suitable for one specific type of substrate – just like a lock that has the right shape for only one specific key. Changing the shape of the active site of an enzyme will cause its reaction to slow down until the shape has changed so much that the substrate no longer fits.
Answer:
A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)
B - Increase(P), Increase(q), Decrease (R)
C - Triple (P) and reduce (q) to one third
Explanation:
<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>
P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.
In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.
Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.
