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Lubov Fominskaja [6]
2 years ago
15

Calculate the mass in grams of each sample.

Chemistry
1 answer:
OlgaM077 [116]2 years ago
4 0

Answer: The answer is A. - 4.88x10^20 H2O2 molecules

Explanation: I hope this helps!

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Calculate how many Calories would be found in a breakfast consisting of the following: two eggs two pieces of toast buttered wit
Lady_Fox [76]
Im assuming around 330?

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Describe the changes in properties (from metals to nonmetals or from nonmetals to metals) as we move (a) down a periodic group a
defon

Answer:

Explanation:

The changes in properties from metals to non-metals on a periodic table can be measured and determined by the metallicity or electropositivity of elements.

Metallicity is a measure of the tendency of atoms of an element to lose electrons.

a.

Down a periodic group, metallicity increases.

b.

Across a period from left to right electropositivity or metallicity decreases.

Metals are found in the left part of periodic table and the most reactive metal sits in the lower left corner. Non-metals are towards the right side of the table.

7 0
2 years ago
Which of these did your answer include? High boiling and melting points: Hydrogen bonds increase the amount of energy required f
scZoUnD [109]

Here we have to get the right answers which include the given phrase.

The correct answers are as following:

High boiling and melting points: Hydrogen bond increase the amount of energy required for phase changes to occur, thereby raising the boiling and melting points.

High specific heat: Hydrogen bond increase the amount of energy required for molecules to increase the speed, thereby raising the specific heat.

High surface tension: Hydrogen bonds produce strong inter molecular attractions, which increase surface tension.

The incorrect answer:

Lower density as a solid than as a liquid: actually, density of solid is more than density of liquid as hydrogen bonds in solid produce strong inter molecular attractions among molecules, which aggregates molecules together, hence volume of associated molecules reduces. Therefore, density of solid is more than that of liquid.

7 0
2 years ago
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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
1. The specific heat capacity of iron is 0.461 J g–1 K–1 and that of titanium is 0.544 J g–1 K–1. A sample consisting of a mixtu
mezya [45]

Answer:

The answer is 80,1 °C

Explanation:

Let´s start from the mass of the sample and the heat capacities:

First of all, we must calculate an average heat capacity. That's because we have a mixture and it is unknown the heat capacity of the whole sample.

The way we should do this calculation is as follows:

(1) H_{average}=Mass Fraction_{first component}* H_{first component}+MassFraction_{secondcomponent}*H_{second component}

For example, the mass fraction of Fe is simply:

(2) MassFraction_{Fe}=\frac{10g Fe}{10g Fe + 10gTi}=0.5

If you combine the equations (1) and (2) you have:

(3) H_{average}=0.5*0.461+0.5*0.544=0.5025\frac{J}{g-K}

Once calculated the average heat capacity we can solve the problem taking into account the corresponding equation:

(4) Q=m*H_{average}*(T_2-T_1)

Remember that:

<em>Q:</em> Heat gained or lost

<em>m</em>: Mass of the sample you want to analize

H_{average} : The value obtained in equation (3)

T_2: Final temperature of the sample

T_1: Initial temperature of the sample

Now we must replace the problem data in equation (4)

Take into account:

  • Heat gained in a system have a positive value
  • Heat lost in a system have a negative value
  • In this problem the sample loses 200 J, for this reason Q=-200J
  • The mass of the whole sample is: 10g of Fe + 10g of Ti = 20g of sample
  • The temperatures must be in absolute units of temperature (these are: rankine or kelvin)
  • The initial temperature of the system is 100°C or 373K

Now we are ready to use equation (4):

(5) -200J=20g*0.5025\frac{J}{g*K} *(T_2-373K)

It is clear that the unknown in equation (5) is T_2

The next step is to calculate T_2. Don't forget the signs; these are important.

Key concept: <u>Since the system is loosing heat, the final temperature of the system ( T_2) should be lower than the initial temperature ( T_1 )</u>

7 0
2 years ago
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