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Roman55 [17]
2 years ago
14

a drop of water weighing 0.48 g condenses on the surface of a 55-g block of aluminum that is initially at 25C. if the heat durin

g condensation goes only toward heating the metal, what is the final temperature in celcius of he metal block? Specific heat capacity of aluminum is 0.903j/g C
Chemistry
1 answer:
Gwar [14]2 years ago
8 0

Mass of water vapor is 0.48 gms.

Weight of the aluminium block is 55 gms.

Heat of vaporization of water at 25 degree Celsius is 44.0 kJ/mol.

The amount of heat given by the condensation is:

q = Heat of vaporization × Mass of vapor / Molar mass of steam

= 44.0 kJ/mol ×0.48 g / 18 g/mol

= 1.173 kJ

Now the final temperature of the metal block is calculated by the formula:

q = m × c × ΔT

q = m × c × (T₂ - T₁)

Here, q is the amount of heat, m is the mass of the metal, c is the specif heat of the metal, T₁ is initial temperature, and T₂ is the final temperature.

Now, substituting the values we get,

1.173 kJ = 55 g × 0.903 J/g. ° C * (T₂ - 25°C)

1.173 × 10³ J = 55 g × 0.903 J/g. degree C × (T₂ - 25°C)

1.173 × 10³ = 49.665 ° C * (T₂ - 25° C)

T₂ = 49° C.

Thus, the final temperature of the metal block is 49° C.

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Why are weight and mass used synonymously on Earth?
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Actually mass and weight are two different things but most people did not understand the difference between them. And they are used synonymously on the earth. Mass is the measure of the amount of matter and weight is the measure of how the force of gravity acts on the mass that is Weight = mass x force of gravity. And they are also directly proportional to each other that is also the reason both terms are used synonymously.

5 0
2 years ago
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What mass (g) of barium iodide is contained in 188 ml of a barium iodide solution that has an iodide ion concentration of 0.532m
Katarina [22]

Answer:

What mass (g) of barium iodide is contained in 188 mL of a barium iodide solution that has an iodide ion concentration of 0.532 M?

A) 19.6

B) 39.1

C) 19,600

D) 39,100

E) 276

The correct answer to the question is

B) 39.1  grams

Explanation:

To solve the question

The molarity ratio is given by

188 ml of 0.532 M solution of iodide.

Therefore we have number of moles = 0.188 × 0.532 M = 0.100016 Moles

To find the mass, we note that the Number of moles = \frac{Mass}{Molar Mass} from which we have

Mass = Number of moles × molar mass

Where the molar mass of Barium Iodide = 391.136 g/mol

= 0.100016 moles ×391.136 g/mol = 39.12 g

8 0
1 year ago
Consider the standard galvanic cell based on the following half-reactions The electrodes in this cell are and . Does the cell po
Free_Kalibri [48]

Question: The question is incomplete and can't be comprehended. See the complete question below and the answer.

Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag 0.80 V Cu2+ + 2 e- → Cu 0.34 V

How would the following changes alter the potential of the cell?

a) Adding Cu2+ ions to the copper half reaction (assuming no volume change).

b) Adding equal amounts of water to both half reactions.

c) Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

d) Adding Ag+ ions to the silver half reaction (assume no volume change)

Explanation:

Nernst equation relates the reduction potential of an electrochemical reaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation.

Reaction under consideration:

Ag+ + e- → Ag 0.80 V

Cu+2 + 2 e- → Cu 0.34 V

Clearly, Ag reduction potential is high and this indicates that it gets reduced readily which leaves Cu to oxidize. Cu+2 ions are products of reaction and Ag+ ions are reactant ions.

Nernst equation : Ecell = E°cell­ – (2.303 RT / n F) log Q    

where                            

             Ecell = actual cell potential

             E°cell­ ­​ = standard cell potential

             R = the universal gas constant = 8.314472(15) J K−1 mol−1

             T = the temperature in kelvins

              n = the number of moles of electrons transferred                                    

 F = the Faraday constant, the number of coulombs per mole of electrons:

  (F = 9.64853399(24)×104 C mol−1)

 Q = [product ion]y / [reactant ion] x

Accordingly when applied to above reaction one will get the following

= E°cell­ – (2.303 x RT / 6 F) log [Cu+2] / [Ag+]

Now the given variables can be studied according to Le Chatelier's principle which states when any system at equilibrium is subjected to change in its concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.

a)        Adding Cu2+ ions to the copper half reaction (assuming no volume change).

Addition of Cu+2 ions increases its concentration and consequently increases the Q value which results in reduction of Ecell. In other words the addition of Cu+2 ions favors the backward reaction to maintain the equilibrium of reaction and hence the forward reaction rate decreases.

b)       Adding equal amounts of water to both half reactions.

Addition of water increases the dilution of the electrochemical cell. For weak electrolytes such as Ag+/ Cu+2 with increase in dilution, the degree of dissociation increases and as a result molar conductance increases.

c)        Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

Based on Le Chatelier's principle when Cu+2 ions amount is decreased by its continuous removal from  the system the forward reaction is favored. As the Cu+2 ions is removed the system attempts to generate more Cu+2 ions to counter the affect of its removal.

d)       Adding Ag+ ions to the silver half reaction (assume no volume change)

Addition of reactant ions, i.e. Ag+ ions, will favour the forward reaction, which results in more product formation.

6 0
2 years ago
The isotope 64Cu has t1/2 of 12.7 hours. If the initial concentration of this isotope in an aqueous solution is 845 ppm, what wi
Kisachek [45]

Answer:

A = 679.2955 ppm

Explanation:

In this case, we already know that 64Cu has a half life of 12.7 hours. The expression to use to calculate the remaining solution is:

A = A₀ e^-kt

This is the expression to use. We have time, A₀, but we do not have k. This value is calculated with the following expression:

k = ln2 / t₁/₂

Replacing the given data we have:

k = ln2 / 12.7

k = 0.0546

Now, let's get the concentration of Cu:

A = 845 e^(-0.0546*4)

A = 845 e^(-0.2183)

A = 845 * 0.8039

A = 679.2955 ppm

This would be the concentration after 4 hours

7 0
1 year ago
During translation, the new polypeptides are often directed to specific parts of the cell by the presence or absence of short se
kotegsom [21]

Answer:

The correct option is b. an amino-terminal signal

Explanation:

A polypeptide that will eventually fold to become an ion channel protein, it means a kind of integral membrane protein, has an amino terminal signal that indicates its delivery to endoplasmic reticulum (ER) and then to the membrane. This type of signal usually consist in a nucleus of 6 to 12 aminoacids and one or more basic aminoacids. Once the polypeptide enters the ER, this signal is removed.

6 0
1 year ago
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