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vodomira [7]
3 years ago
15

A sample from one of Earth's oceans has a salinity of 34. What is the concentration of dissolved salts in this sample of seawate

r when expressed in ppmppm
Chemistry
1 answer:
11Alexandr11 [23.1K]3 years ago
3 0

Answer:

Concentration of dissolved salts = 34,038.76 ppm

Explanation:

Given:

Salinity of ocean water = 34

Find:

Concentration of dissolved salts

Computation:

Salinity of ocean water = 34 g/l

1g/l = 1001.14 ppm

Concentration of dissolved salts = 1001.14 ppm x 34

Concentration of dissolved salts = 34,038.76 ppm

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Mendeleev used his periodic table of elements to
pishuonlain [190]
<span>Mendeleev realized that the physical and chemical properties of elements were related to their atomic mass in a 'periodic' way, and arranged them so that groups of elements with similar properties fell into vertical columns in his table.
</span>Mendeleev rewrote each edition of Principles of Chemistry<span>, including all new scientific data-particularly confirmations of the periodic law-and reanalyzing difficulties that had arisen to hinder its confirmation (inert gases, radioactivity, radioactive and </span>rare-earth elements<span>)
</span>he wanted to prove that he was right.
7 0
2 years ago
Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston-cylinder assembly from 2 bar, 280 K to 20 bar
kaheart [24]

This question is incomplete, the complete question is;

Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston-cylinder assembly from 2 bar, 280 K to 20 bar, 520 K. If the carbon dioxide behaves as an ideal gas, determine the amount of entropy produced, in kJ/K. Assuming;

a) constant specific heats Cp = 0.939 kJ/Kg K

b) variable specific heats

Answer:

a) the amount of entropy produced is 0.731599 kJ/K

b) the amount of entropy produced is 0.69845 kJ/K

Explanation:

Given the data in the question;

5 kg of carbon dioxide (CO₂) gas undergoes a process in a well-insulated piston-cylinder assembly.

m = 5 kg

Molar mass M = 44.01 g/mol

P₁ = 2 bar, P₂ = 20

T₁ = 280 K, P₂ = 520 K

Since its insulated { q = 0 } ( kinetic and potential energy effects = 0 )

Now,

a) the amount of entropy produced, in kJ/K, Assuming constant specific heats with Cp = 0.939 kJ/Kg K

S_{Generation = m × ((Cp × In( T₂/T₁) - R × In( P₂/p₁ ))

we substitute

S_{Generation = 5 × (( 0.939  × In( 520/280) - 0.1889 × In( 20/2 ))

= 5 × ( 0.5812778 - 0.434958 )

= 5 × 0.1463198

= 0.731599 kJ/K

Therefore, the amount of entropy produced is 0.731599 kJ/K

b) the amount of entropy produced, in kJ/K, Assuming variable specific heats.

Now, from  Table A-23: Ideal Gas Properties of Selected Gases;

T₁,T₂ : s₁⁰ = 211.376 kJ/kmol-K, s₂⁰ = 236.575 kJ/kmol-K

now, s₁ = s₁⁰ / M and s₂ = s₂⁰ / M

we substitute

s₁ = s₁⁰ / M = 211.376 / 44.01  = 4.8029 kJ/kg

s₂ = s₂⁰ / M = 236.575 / 44.01 = 5.37548 kJ/kg

S_{Generation = m × (( s₂ - s₁ ) - R × In( p₂ / p₁ ))

we substitute

S_{Generation = 5 × (( 5.37548 - 4.8029  ) - 0.1880 × In( 20 / 2 ))

= 5 × ( 0.57258 - 0.432885997 )

= 5 × 0.13969

= 0.69845 kJ/K

Therefore, the amount of entropy produced is 0.69845 kJ/K

5 0
2 years ago
PLS HELP ASAP!!!! Describe metallic bonds, include an explanation of their valence electrons. How this related to the properties
Aloiza [94]

If you look it up it will give you plenty of information. This is what I found:

The valence electrons of metals move freely in this way because metals have relatively low electronegativity, or attraction to electrons. The positive metal ions form a lattice-like structure held together by all the metallic bonds. ... When nonmetals bond together, the atoms share valence electrons and do not become ions

https://www.ck12.org/c/physical-science/metallic-bond/lesson/Metallic-Bonding-MS-PS/

5 0
3 years ago
This is agree or disagree I need help ASAP!!
PIT_PIT [208]
All of them are agree :)
8 0
2 years ago
Read 2 more answers
You work for a cutlery manufacturer who wants to electrolytically precipitate 0.500 g of silver onto each piece of a batch of 25
ICE Princess25 [194]

Answer:

373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.

Explanation:

Mass of silver to be precipitated on ecah spoon = 0.500 g

Number of silver spoons = 250

Total mass of silver = 250 × 0.500 g = 125 g

Moles (n)=Molarity(M)\times Volume (L)

Moles of AgCN = n  = \frac{125 g}{134 g/mol}=0.9328 mol

Volume of AgCN solution =V

Molarity of the AgCN = 2.50 M

V=\frac{0.9328 mol}{2.50 M}=0.3731 L=373.1 mL

(1 L = 1000 mL)

373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.

6 0
3 years ago
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