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3 years ago

12

I just need some help getting the answers for this portfolio, but if someone could not only give the answers, but also explain h

ow to get those answers so I better understand how to do this myself... It would be greatly appreciated.

1 answer:

vova2212 [387]3 years ago

8 0

Answer:

The balanced chemical reaction has equal number of atoms in reactant side as well as product side.

Explanation:

The balanced chemical reactions is as follows.

1.

$Zn+2HCl\rightarrow ZnCl_{2}+H_{2}$

2.

$4Fe+3O_{2}\rightarrow 2Fe_{2}O_{3}$

3.

$2H_{2}O_{2}\rightarrow 2H_{2}O +O_{2}$

4.

$NaHCO_{3}+HC_{2}H_{3}O_{2} \rightarrow NaC_{2}H_{3}O_{2}+H_{2}CO_{3}$

$H_{2}CO_{3}\rightarrow H_{2}O+CO_{2}$

$NaHCO_{3}+HC_{2}H_{3}O_{2}\rightarrow H_{2}O+CO_{2}+NaC_{2}H_{3}O_{2}$

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Which statement defines reduction potential when considering a pair of half-cell reactions?

Ivahew [28]

The half reaction with the the greater SRP has a greater tendency to gain electrons is the definition of reduction potential when considering a pair of half cell reactions.This reduction potential is measured against hydrogen electrode which is standard electrode.

6 0

3 years ago

Suppose that 100 grams of water at 50.0°C is placed in contact with 200 grams of iron at 30.0°C. The final

wel

Answer:

The answer would be 1.5 kJ.

Explanation:

When you use the equation q = m x c x ∆T you will be able to find the energy gained or lost. The data for the water in this case is just there to distract you so ignore it. :D

4 0

2 years ago

At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T

Anarel [89]

<u>Answer:</u> The freezing point of solution is -0.974°C

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M$

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

$1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

$0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (glucose) = 89.18 g

$M_{solute}$ = Molar mass of solute (glucose) = 180.16 g/mol

$W_{solvent}$ = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC$

Hence, the freezing point of solution is -0.974°C

8 0

3 years ago

0.0200 moles of a compound is found to have a mass of 1.64 g. Find the formula mass of the compound

KatRina [158]

Answer: 82.0 g/mole

Explanation:

Use the units to see that if we divide 1.64 grams by 0.0200 moles, we'll get a number that is grams/mole, the definition of formula mass.

1.64/0.0200 = 82.0 g/mole (3 sig figs)

We can't tell from this alone what the molecular formula might be, but C6H10 (cyclohexene) comes close (82.1 grams/mole).

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2 years ago

How do scientists distinguish between an element and a compound? List two elements and two compounds.

Vsevolod [243]

The difference between an element and a compound is that an element is composed of only one kind of atom while a compound is composed of at least 2 kinds of atoms. The properties of an element and the compound the atoms has to form are different physically and chemically.Example of element are O and H, while compound examples are H2O and H2O2.

7 0

3 years ago