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harkovskaia [24]
2 years ago
8

An earthquake on the ocean floor produced a giant wave called a tsunami. The tsunami traveled through the ocean and hit a remote

island, causing a lot of damage. Is the water that hit the island the same water that was above the earthquake on the ocean floor?
Chemistry
1 answer:
Svetlanka [38]2 years ago
8 0

Answer:

Yes

Explanation:

What cause an earthquake is when the earth plates shift and if theirs a drop in the tentonic plates a ripple effect like when you drop something in water will occur. The plates shift down the water in which the plate shift down the water will go in that direction due to gravity, but instead of equalizing the water will pick up some speed and velocity and begin to form a wave now. When an tsunami happens you know it coming cause the water moves back cause the water is picking up to much speed and due to cohesion its moves along with the move water and builds up. Creating a massive tidal wave known as tsunamis.

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During the lab, you will have access to a range of acids and bases as well as universal pH indicator paper. Think about how you
kobusy [5.1K]

Answer:

Explanation:bxbxbd

6 0
2 years ago
Read 2 more answers
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
A sample of copper with a mass of 63.5g contains 6.02 x10^23 atoms calculate the mass of an average copper atom
m_a_m_a [10]

Answer:

The mass of an average copper atom is 1.0548\times 10^{-22}\ g

Explanation:

Given:

The total mass of copper atoms, m = 63.5\ g

Number of atoms, N=6.02\times 10^{23}

Now, we are asked to find the mass of 1 copper atom.

We use unitary method to find the mass of 1 copper atom.

Mass of N atoms = m

∴ Mass of 1 atom = \frac{m}{N}

Plug in 63.5 for 'm', 6.02\times 10^{23} for 'N' and simply.

Mass of 1 atom = \dfrac{63.5}{6.02\times 10^{23}}=1.0548\times 10^{-22}\ g

Therefore, the mass of an average copper atom is 1.0548\times 10^{-22}\ g

5 0
2 years ago
H2A and BOH are acid and base and they react according to the following balanced equation: H2A(aq) + 2 BOH(aq) → B2A(aq) + 2 H2O
Rudiy27

Answer:

ΔH=15000 J  =  15KJ

Explanation:

In this exercise  you have find the enthalpy of reaction this is the difference between enthalpy of reactans and products,

For the following equation

H2A(aq) + 2 BOH(aq) → B2A(aq) + 2 H2O(l)

We know that 0.20 moles of BOH reacted with excess amount of H2A solution and 1500. J

so,

(2mol/0,2mol)*1500J=15000J

for de reactions exothermics tha enthalpy is negative so:

ΔH=15000 J  =  15KJ

3 0
2 years ago
. If one mole each of CH4, NH3, H2S, and CO2 is added to 1 liter of water in a flask (1 liter of water = 55.5 moles of H2O), how
denpristay [2]
1 litre of water is = 55.5 moles of water.
water is H2O
so, in water:
moles of oxygen = 55.5
moles of hydrogen = 2 x 55.5 = 111

Now, 1 mole each of <span>CH4, NH3, H2S, and CO2 are added:
For CH4: 
moles of C = 1
moles of H = 4 x 1 = 4

For NH3:
moles of N = 1
moles of H = 3 x 1 = 3

For H2S:
moles of H = 2 x 1 = 2
</span>moles of S = 1
<span>
For CO2:
</span>moles of C = 1
moles of  = 2 x 1 = 2
<span>
Now, add the total moles of each atom:
Hydrogen = 111 + 4 + 3 +1 = 119 moles
Oxygen = 55.5 + 2 = 57.5
Carbon = 1+1 = 2
Sulfur = 1
nitrogen = 1

</span>
6 0
2 years ago
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