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Veseljchak [2.6K]
3 years ago
7

An increase in ocean salinity can increase ____, creating a current.

Chemistry
2 answers:
g100num [7]3 years ago
8 0
B density I think.. sry if I’m wrong
horrorfan [7]3 years ago
5 0

Answer:

B density

Explanation:

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Calculate the hydrogen-ion concentration [H+] for the aqueous solution in which [OH–] is 1 x 10–11 mol/L. Is this solution acidi
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(H+)(OH-) = Kw
kw=  1 x10^-14
OH-=  1   x10 ^-11
(H+)=  KW / OH-

concentration   of H+  = (1x10^-14) /.(1  x 10 ^-11)   =  1  x10  ^-3

Ph=  -log (H+)

PH=-log (  1  x  10  ^-3)  =  3  therefore  the    solution  is  acidic  since  the  PH   less than  7

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Calculate the value of AE for a system that loses 50 J of heat and has 150 J of work Performed on it by the surroundings.
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100 j

Explanation:

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4. On the island of Hawaii, Keanu notices that the sand on the beach is black, the same color as the
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The ricks were degraded over time by wind and water and were crushed down into tiny, smaller rocks, which is sand.

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Products or solutions found at home or in store and their characteristics
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3 years ago
Phosgene, COCl2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide w
jek_recluse [69]

The question is incomplete, here is the complete question:

Phosgene, COCl_2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide with chlorine:  

CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

The value of K_c for this reaction is 5.79 at 570 K. What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which p_{CO}=p_{Cl_2}=0.265atm and p_{COCl_2}=0.000atm ?

<u>Answer:</u> The equilibrium partial pressure of CO, Cl_2\text{ and }COCl_2 is 0.257 atm, 0.257 atm and 0.008 atm respectively.

<u>Explanation:</u>

The relation of K_c\text{ and }K_p is given by:

K_p=K_c(RT)^{\Delta n_g}

K_p = Equilibrium constant in terms of partial pressure

K_c = Equilibrium constant in terms of concentration = 5.79

\Delta n_g = Difference between gaseous moles on product side and reactant side = n_{g,p}-n_{g,r}=1-2=-1

R = Gas constant = 0.0821\text{ L. atm }mol^{-1}K^{-1}

T = Temperature = 570 K

Putting values in above equation, we get:

K_p=5.79\times (0.0821\times 570)^{-1}\\\\K_p=0.124

We are given:

Initial partial pressure of CO = 0.265 atm

Initial partial pressure of chlorine gas = 0.265 atm

Initial partial pressure of phosgene = 0.00 atm

The given chemical equation follows:

                      CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

<u>Initial:</u>            0.265      0.265

<u>At eqllm:</u>        0.265-x    0.265-x        x

The expression of K_p for above equation follows:

K_p=\frac{p_{COCl_2}}{p_{CO}\times p_{Cl_2}}

Putting values in above equation, we get:

0.124=\frac{x}{(0.265-x)\times (0.265-x)}\\\\x=0.0082,8.59

Neglecting the value of x = 8.59 because equilibrium partial pressure cannot be greater than initial pressure

So, the equilibrium partial pressure of CO = (0.265-x)=(0.265-0.008)=0.257atm

The equilibrium partial pressure of Cl_2=(0.265-x)=(0.265-0.008)=0.257atm

The equilibrium partial pressure of COCl_2=x=0.008atm

Hence, the equilibrium partial pressure of CO, Cl_2\text{ and }COCl_2 is 0.257 atm, 0.257 atm and 0.008 atm respectively.

6 0
3 years ago
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