The pressure of sulfur dioxide in a container is 159 kpa. what is this pressure in atmospheres?

The new boiling point is 101.02°C. Suppose that you add another 2.0 mol of sucrose to this solution. What do you predict the new

solniwko [45]

Mad Lad, thanks for the answer :)

7 0

3 years ago

Read 2 more answers

Choose the correct option for the redox reaction below : 8H+(aq) + MnO4-(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) 4H2O(l) H+ (oxid

hodyreva [135]

Explanation:

8H+(aq) + MnO4-(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

To determine if a substance was reduced or oxidized, the oxidation number at the reactant and product side is compared.

When oxidation occurs, an element loses electrons and its oxidation number increases (becomes more positive). When reduction occurs, an element gains electrons and its oxidation number decreases or is reduced (becomes more negative).

H+ (oxidized or reduced) Answer 1 Choose...

There is an increase in oxidation number from 0 to 1. This means H+ was oxidized.

MnO4- (oxidizing or reducing agent) Answer 2 Choose...

In chemistry, an oxidizing agent is a substance that has the ability to oxidize other substances — in other words to accept their electrons

MnO4- is a strong oxidizing agent.

Fe2+ (oxidizing or reducing agent) Answer 3 Choose...

A reducing agent means it will itself get oxidized. In this reaction, Fe2+ is the reducing agent.

Fe2+ (oxidized or reduced) Answer 4 Choose...

Oxidized

O in MnO4- (oxidized or reduced) Answer 5 Choose...

There is no change in oxidation number of , so it was neither reduced nor was it oxidized.

Mn in MnO4- (oxidized or reduced) Answer 6 Choose...

There is a decreaase in oxidation number from +7 to +2. This means reduction.

6 0

3 years ago

What volume of 0.160 m li2s solution is required to completely react with 130 ml of 0.160 m co no3 2?

Over [174]

The volume of 0.160 m Li2S solution required to completely react with 130 ml of 0.160 CO(NO3)2 is calculated as below

write the reacting equation

Co(NO3)2 + Li2S = 2LiNO3 + COS

find the moles of CO(NO3)2 = molarity x volume

= 130 ml x 0.160=20.8 moles

since the reacting moles between CO(NO3)2 to LiS is 1:1 the moles of LiS is also 20.8 moles

volume of Lis is therefore = moles of Lis/ molarity of LiS

= 20.8/0.160 = 130 Ml

3 0

3 years ago

What are devices that move atomic nuclei at extremely high speed

Temka [501]

The answer is particle accelerators.

7 0

3 years ago

Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th

Tpy6a [65]

$pH = 13.5$

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

$\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}$

The mixture would contain

$0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol}$ of $\text{OH}^{-}$ and
$0.1 \times 0.5 = 0.05 \; \text{mol}$ of $\text{Ac}^{-}$

if $\text{Ac}^{-}$ undergoes no hydrolysis; the solution is of volume $0.1 + 0.4 = 0.5 \; \text{L}$ after the mixing. The two species would thus be of concentration $0.30 \; \text{mol} \cdot \text{L}^{-1}$ and $0.10 \; \text{mol} \cdot \text{L}^{-1}$ , respectively.

Construct a RICE table for the hydrolysis of $\text{Ac}^{-}$ under a basic aqueous environment (with a negligible hydronium concentration.)

$\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}$

The question supplied the <em>acid</em> dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant $pK_b$ for its conjugate base, $\text{Ac}^{-}$ . The following relationship relates the two quantities:

$pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})$

... where the water self-ionization constant $pK_w \approx 14$ under standard conditions. Thus $pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3$ . By the definition of $pK_b$ :

$[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}$

$x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}$

$x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}$

$[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}$

$pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5$

6 0

3 years ago