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3 years ago

Aluminum has a density of 2.70 g/mL. Calculate the mass (in grams) of a piece of aluminum having a volume of 405 mL .

1 answer:

7 0

In this question, you are given the density (2.7g/ml) and the volume of the aluminum. You are asked how much the volume is. To answer this question, you need to convert the volume into mass. Since the unit used is already same, you don't need to convert it into another unit. The calculation would be:

mass= volume / density
mass= 405ml / (2.70 g/ml)
mass= 150grams

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You mix 125 mL of 0.170 M with 50.0 mL of 0.425 M in a coffee-cup calorimeter, and the temperature of both solutions rises from

kondaur [170]

Here is the correct question

You mix 125 mL of 0.170 M CsOH with 50.0 mL of 0.425 M HF in a coffee-cup calorimeter, and the temperature of both solutions rises from 20.20 °C before mixing to 22.17 °C after the reaction. What is the enthalpy of reaction per mole of ? Assume the densities of the solutions are all 1.00 g/mL, and the specific heat capacities of the solutions are 4.2 J/g · K. Enthalpy of reaction = kJ/mol

Answer:

75.059 kJ/mol

Explanation:

The formula for calculating density is:

$density = \frac{mass}{volume}\\$

Making mass the subject of the formula; we have :

mass = density × volume

which can be rewritten as:

mass of the solution = density × volume of the solution

= 1.00 g/mL × (125+ 50 ) mL

= 175 g

Specific heat capacity = 4.2 J/g.K

∴ the energy absorbed is = mcΔT

= 175 × 4.2 × (22.17 - 20.00) ° C

= 1594.95 J

= 1.595 J

number of moles of CsOH = $\frac{125}{1000} *100$

= 0.2125 mole

Therefore; the enthalpy of the reaction = $\frac{Energy \ absorbed }{number \ of \ moles}$

= $\frac{1.595}{0.02125}$

= 75.059 kJ/mol

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3 years ago

As the human population grows, some minerals in everyday products could

mylen [45]

Answer: explore uses of more plentiful minerals

Explanation: ap!x

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3 years ago

Calculate # of atoms in 46g of Na

makvit [3.9K]

21Explanation:BEACUSE I CAN

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3 years ago

2.61 kilograms of water in a container have a pressure of 200 kPa and temperature of 200°C . What is the volume of this containe

Arisa [49]

Answer: The volume of the container is $2.8497m^3$

Explanation:

To calculate the volume of water, we use the equation given by ideal gas, which is:

$PV=nRT$

or,

$PV=\frac{m}{M}RT$

where,

P = pressure of container = 200 kPa

V = volume of container = ? L

m = Given mass of water = 2.61 kg = 2610 g (Conversion factor: 1kg = 1000 g)

M = Molar mass of water = 18 g/mol

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of container = $200^oC=[200+273]K=473K$

Putting values in above equation, we get:

$200kPa\times V=\frac{2610g}{18g/mol}\times 8.31\text{L kPa }\times 473K\\\\V=2849.7L$

Converting this into cubic meter, we use the conversion factor:

$1m^3=1000L$

So, $\Rightarrow \frac{1m^3}{1000L}\times 2849.7L$

$\Rightarrow 2.8497m^3$

Hence, the volume of the container is $2.8497m^3$

6 0

3 years ago

In thermodynamics, we determine the spontaneity of a reaction by the sign of ΔG. In electrochemistry, spontaneity is determined

faltersainse [42]

Answer:

For A: The standard cell potential of the reaction is 4.4 V

For B: The standard Gibbs free energy of the reaction is $-8.50\times 10^5J$

For C: The reaction is spontaneous as written.

Explanation:

For A:

The given chemical reaction follows:

$2Li(s)+Cl_2(g)\rightarrow 2Li^+(aq.)+2Cl^-(aq.)$

The given half reaction follows:

Oxidation half reaction: $Li(s)\rightarrow Li^+(aq.)+e^-;E^o_{Li^+/Li}=-3.04V$ ( × 2)

Reduction half reaction: $Cl_2(g)+2e^-\rightarrow 2Cl^-(aq.);E^o_{Cl_2/2Cl^-}=+1.36V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

Here, chlorine will undergo reduction reaction will get reduced.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=1.36-(-3.04)=4.4V$

Hence, the standard cell potential of the reaction is 4.4 V

For B:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

n = number of electrons transferred = $2mol\text{ e}^-$

F = Faradays constant = $96500J/V.mol\text{ e}^-$

$E^o_{cell}$ = standard cell potential = 4.4 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 4.4=-849200J=-8.50\times 10^5J$

Hence, the standard Gibbs free energy of the reaction is $-8.50\times 10^5J$

For C:

For a reaction to be spontaneous, the standard Gibbs free energy change of the reaction must be negative.

From above, the standard Gibbs free energy change of the reaction is coming out to be negative.

Hence, the reaction is spontaneous as written.

7 0

4 years ago