Answer:
0.319 L
Explanation:
M(MgSO4) = 120 g/mol
46.1 g * 1 mol/120 g = 0.384 mol MgSO4
0.384 mol * 1 L/1.20 mol = 0.319 L
Answer:
A.)
Explanation:
Just took this test and got the question right!
Answer:
a. 2nd order reaction.
b. The first step is the slow step.
Explanation:
r = k[NO][Cl₂]
a. The reaction is first-order in [NO] and first-order in [Cl₂], so it is second-order overall.
b. The first step is the slow step, because it predicts the correct rate law.
c. is wrong. Doubling [NO] would double the rate, because the reaction is first-order in [NO].
d. is wrong. Cutting [Cl₂] in half would halve the rate, because the reaction is first-order in [Cl₂].
e. is wrong. The molecularity is two, because two particles are colliding.
f. is wrong. Both steps are bimolecular.
The electrode A is a cathode. This is because it is negatively charged electrode and it attracts cations, or we can say it is positively charged.
The reaction occurs at the cathode
Zn2+ +2e⇒Zn
Anode reaction is
2CL- ⇒CL2+2e-
The overall reaction will be Zncl2(l) which completely disassociates then it is zn∧2+(l)+2cl∧-(l).
Answer:
7.28 moles Ag°
Explanation:
Cu° + 2 AgNO₃ => Cu(NO₃)₂ + 2Ag°
Given 7.28 moles 7.28 moles
To determine limiting reactant, divide the mole values by the respective coefficient of balanced equation. The resulting smallest value is the limiting reactant. Note: this is a short cut method for determining limiting reactant only. Once the limiting reactant is determined one must use the given mole values of the limiting reactant to solve problem. That is ...
Limiting reactant determination:
Cu° + 2 AgNO₃ => Cu(NO₃)₂ + 2Ag°
Cu: 7.28 / 1 = 7.28
AgNO₃ : 7.28 / 2 = 3.64 => Limiting Reactant is AgNO₃
Solving Problem depends on AgNO₃; Cu will be in excess.
Since coefficients of AgNO₃ & Ag° are equal, then the moles AgNO₃ used equals moles Ag° produced and is therefore 7.28 moles Ag°.
