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2 years ago

How might a recent college graduate’s investment portfolio differ from someone who is nearing retirement?

Chemistry

2 answers:

Arisa [49]2 years ago

7 0

Answer:

The recent graduate invests taking risks and betting on profitability.

The retired man protects the estate by sacrificing the gain.

Explanation:

Pepsi [2]2 years ago

5 0

Older people had no money to buy stuff

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Analyze and solve this partially completed galvanic cell puzzle. There are 4 electrodes each identified by a letter of the alpha

klasskru [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is $E_{cell}__{AC}} = 0.94$

Explanation:

From the question we are told that

the cell voltage for AD is $E_{cell}__{AD}} = 1.56V$

From the data give we can see that

$E_{cell}__{AD}} - E_{cell}__{BD}} = E_{cell}__{AB}}$

i.e $1.56 - 1.53 = 0.03$

In the same way we can say that

$E_{cell}__{AD}}-E_{cell}__{CD}} = E_{cell}__{AC}}$

=> $E_{cell}__{AC}}=1.56- 0.62$

$E_{cell}__{AC}} = 0.94$

5 0

2 years ago

The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t

Basile [38]

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

$2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?$

We have :

Enthalpy changes of formation of following s:

$\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol$

$\Delta H_{f,CuO}=-157.3 kJ/mol$

$\Delta H_{f,NO_2}= 33.2 kJ/mol$

$\Delta H_{f,O_2}= 0 kJ/mol$ (standard state)

$\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]$

The equation for the enthalpy change of the given reaction is:

$\Delta H_{rxn}$ =

$=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})$

$\Delta H_{rxn}$ =

$(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)$

$\Delta H_{rxn}=424 kJ$

The enthalpy change for the given reaction is 424 kJ.

6 0

2 years ago

A hydrogen filled balloon has a bolume of 8.3 L at 36 C and 751 torr. How many moles of hydrogen are inside the balloon?

alina1380 [7]

Answer:

0.33 mol

Explanation:

Given data:

Volume of balloon = 8.3 L

Temperature = 36°C

Pressure = 751 torr

Number of moles of hydrogen = ?

Solution:

Temperature = 36°C (27 +273 = 300 K)

Pressure = 751 torr (751/760= 0.988 atm)

Formula:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

PV = nRT

0.988 atm × 8.3 L = n × 0.0821 atm.L/ mol.K × 300 K

8.2 atm.L = n × 24.63 atm.L/ mol

n = 8.2 atm.L / 24.63 atm.L/ mol

n = 0.33 mol

8 0

2 years ago

When 13.6 g of calcium chloride, CaCl2, was dissolved in 100.0 mL of water in a coffee cup calorimeter, the temperature rose fro

DanielleElmas [232]

Answer:

THE ENTHALPY OF SOLUTION IS 3153.43 J/MOL OR 3.15 KJ/MOL.

Explanation:

1. write out the variables given:

Mass of Calcium chloride = 13.6 g

Change in temperature = 31.75°C - 25.00°C = 6.75 °C

Density of the solution = 1.000 g/mL

Volume = 100.0 mL = 100.0 mL

Specific heat of water = 4.184 J/g °C

Mass of the water = unknown

2. calculate the mass of waterinvolved:

We must first calculate the mass of water in the bomb calorimeter

Mass = density * volume

Mass = 1.000 * 100

Mass = 0.01 g

3. calculate the quantity of heat evolved:

Next is to calculate the quantity of heat evolved from the reaction

Heat = mass * specific heat of water * change in temperature

Heat = mass of water * specific heat *change in temperature

Heat = 13.6 g * 4.184 * 6.75

Heat = 13.6 g * 4.184 J/g °C * 6.75 °C

Heat = 384.09 J

Hence, 384.09J is the quantity of heat involved in the reaction of 13.6 g of calcium chloride in the calorimeter.

4. calculate the molar mass of CaCl2:

Next is to calculate the molar mas of CaCl2

Molar mass = ( 40 + 35.5 *2) = 111 g/mol

The number of moles of 13.6 g of CaCl2 is then:

Number of moles of CaCl2 = mass / molar mass

Number of moles = 13.6 g / 111 g/mol

Number of moles = 0.1225 mol

So 384.09 J of heat was involved in the reaction of 1.6 g of CaCl2 in a calorimter which translates to 0.1225 mol of CaCl2..

5. Calculate the enthalpy of solution in kJ/mol:

If 1 mole of CaCl2 is involved, the heat evolved is therefore:

Heat per mole = 384.09 J / 0.1225 mol

Heat = 3 135.43 J/mol

The enthalpy of solution is therefore 3153.43 J/mol or 3.15 kJ/mol.

5 0

2 years ago

Which three temperature readings all mean the same thing? Question options: 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit

shtirl [24]

273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit

4 0

2 years ago

Read 2 more answers