The titration of 25.0 mL of an unknown concentration H2SO_4 solution requires 83

Question

The titration of 25.0 mL of an unknown concentration H2SO_4 solution requires 83.6 mL of 0.12 M LiOH solution. What is the concentration of the H_2SO_4 solution (in M)? 0.20 M 0.40 M 0.10M 0.36 M 0.25M Determine the oxidation state of P in PO_3^3-. What volume of a 0.540 M NaOH solution contains 11.5 2 of NaOH? What element is undergoing oxidation (if any) in the following reaction? This is not an oxidation-reduction reaction. A 0.3340 g sample of an unknown halogen occupies 109.0 mL at 398.0 K and 1.412 atm. What is the identity of the halogen? Determine the volume of O_2 (at STP) formed when 50.0 g of KClO_3 decomposes according to the following reaction. The molar mass for KCIO_3 is 122.55 g/mol.

Explanation / Answer

18.

H2SO4 + 2 LiOH Li2SO4 + 2 H2O

(83.6 mL) x (0.12 M LiOH) x (1 mol H2SO4 / 2 mol LiOH) / (25.0 mL H2SO4) = 0.20 M H2SO4

so answer is A 0.20 M

19.

The oxidation state of O is -2.
Let the oxidation state of P be x.

The net charge on the ion is -3.
The algebraic sum of the oxidation states of P and O should result in the overall charge on the given species.
Thus, x + 3(-2) = -3
x - 6 = -3
x = +3

Thus, the oxidation state of P is +3 in (PO3)3-.

so answer is A +3

20.

Find the number of moles of given NaOH:
11.5 g NaOH * (1 mole NaOH / 39.9971 g NaOH)
= 0.287 moles NaOH

Now put the numbers into this formula:
molarity = moles NaOH / L
0.540 M = 0.287 moles NaOH / L
= 0.532 L

so answer is B.0.532 L

21.

Zn is going under oxidation its state is going to change from 0 to +2

so answer is A. Zn

22

.PV=nRT
solve for moles (n)
n = PV/RT = 1.412 atm x 0.109 L / (0.082057 x 398 K) = 0.004706 mole
0.334 g / 0.004706 mole = 70.97 g/mole
Cl2 (70.9 g/mole)

so answer is C) Cl2

23.

2 KClO3(s) 2 KCl(s) + 3 O2(g)

50.0g KClO3 x (1 mole KClO3 / 122.6g KClO3) x (3 moles O2 / 2 moles KClO3) x (22.4L / mole) = 13.7L

so answer is D)  13.7L

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