Which of these is NOT a property of a wave?

6 Fe2+ (aq) + Cr2O72− (aq) + 14 H+ (aq) → 6 Fe3+ (aq) + 2 Cr3+ (aq) + 7 H2O (aq) If the titration of 23 mL of an iron(II) soluti

ratelena [41]

Answer:

1.047 M

Explanation:

The given reaction:

$6Fe^{2+}_{(aq)}+Cr_2O_7^{2-}_{(aq)}+14H^+_{(aq)}\rightarrow 6Fe^{3+}_{(aq)}+2Cr^{3+}_{(aq)}+7H_2O_{(aq)}$

For dichromate :

Molarity = 0.254 M

Volume = 15.8 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 15.8 ×10⁻³ L

Thus, moles of dichromate :

$Moles=0.254 \times {15.8\times 10^{-3}}\ moles$

Moles of dichromate = 0.0040132 moles

1 mole of dichromate react with 6 moles of iron(II) solution

Thus,

0.0040132 moles of dichromate react with 6 × 0.0040132 moles of iron(II) solution

Moles of iron(II) solution = 0.02408 moles

Volume = 23 mL = 0.023 L

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

<u>Molarity = 0.02408 / 0.023 = 1.047 M</u>

7 0

2 years ago

What type of flower is this? This is for a science project.

kenny6666 [7]

I believe that is a plant called Wisteria.

4 0

3 years ago

Read 2 more answers

What effect does increasing the concentration of a dissolved solute have on each of the colligative properties?

Igoryamba

Answer:

If we increase the concentration of a dissolved solute, the solution would have a vapor pressure so much low, the boiling temperature for the solution will be so high, freezing point for the solution will be so much low and the osmotic pressure will be higher.

Colligative properties always depends on dissolved particles (solute)

Explanation:

These are the colligative properties

- Vapor pressure lowering

ΔP = P° . Xm

Vapor pressure of pure solvent - Vapor pressure of solution.

If we add more solute, it would raise the Xm, so the solution would have a vapor pressure so much low.

Vapor pressure pure solvent - Vapor pressure solution ↑ = P° . Xm ↑

- Boiling point elevation

ΔT = Kb . m

When we add more solute, we are increasing the molality.

↑T° boiling of solution - T° boiling pure solvent = Kf . m ↑

Boiling temperature for the solution will be so high.

- Freezing point depression

When we add more solute, we are increasing the molality.

ΔT = Kf . m

T° fussion of pure solvent - ↓T° fussion of solution = Kf . m↑

Freezing point for the solution will be so much low.

- Osmotic pressure

π = M . R . T

When we add solute, molarity is increasing. Therefore the osmotic pressure will be higher.

π↑ = M↑ . R . T

7 0

3 years ago

Here is a more complex redox reaction involving the dichromate ion in acidic solution: 3N O 2 − + 8H + + C r 2 O 7 2− → 3N O 3 −

Fantom [35]

Answer:

NO2- is the reducing agent.

Cr2O7_2- is the oxidizing agent.

H+ is neither

Explanation:

Reduction is the gain in electron. A chemical specie that undergoes reduction is called the oxidizing agent.

Oxidation is simply the loss in electrons. A chemical specie that undergoes oxidation is called the reducing agent.

Let us look at the species.

The first specie is the NO2-. In this specie, the oxidation number of nitrogen changed from +3 to +5 in NO3-. Thus we can see that there is more loss of electron to have caused an increase in the oxidation number positively. This shows an oxidation. Hence, NO2- is the reducing agent.

Let us look at the chromium. We can see that the oxidation number of chromium changed from +7 to +3.

Now we can see that it is a decrease and hence, it is a gain of electron and thus it is reduction. This means the first chromium specie is the oxidizing agent.

The hydrogen ion is simply placed there to balance the ions and hence it is neither the oxidizing nor the reducing agent.

4 0

3 years ago

11.0 L of hydrogen and 5.52 L of oxygen are exploded together in a reaction tube. What volume of water vapor was formed, at STP?

Marysya12 [62]

Answer:

11.0 L

Explanation:

The equation for this reaction is given as;

2H2 + O2 --> 2H2O

2 mol of H2 reacts with 1 mol of O2 to form 2 mol of H2O

At STP;

1 mol = 22.4 L

This means;

44.8 L of H2 reacts with 22.4 L of O2 to form 44.8 L of H2O

In this reaction, the limiting reactant is H2 as O2 is in excess.

The relationship between H2 and H2O;

44.8 L = 44.8 L

11.0 L would produce x

Solving for x;

x = 11 * 44.8 / 44.8

x = 11.0 L

4 0

3 years ago