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Ilia_Sergeevich [38]
2 years ago
14

Fdsddddgdgdgdgsdggd.

Chemistry
1 answer:
larisa [96]2 years ago
4 0
Fdsddadgdsadgsd:
Fasdadasdfadsdfasdf asfsadafasdfasdffdaddad dads fdfgdfgfdg

Fdgdfgdfgdgfggggfg gdfgdfgd:
Fgdfgdfgdfgrdfhdf hdfgdfhdfh == fhdfgdfh /fhdgdg .
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rosijanka [135]
A generator because they connect each other in physical attraction
7 0
2 years ago
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Calculate the volume 3.00 moles of a gas will occupy at 24.0˚C and 1.003 atm. *
vitfil [10]

Answer: 72.93 litres

Explanation:

Given that:

Volume of gas (V) = ?

Temperature (T) = 24.0°C

Convert 24.0°C to Kelvin by adding 273

(24.0°C + 273 = 297K)

Pressure (P) = 1.003 atm

Number of moles (n) = 3 moles

Molar gas constant (R) is a constant with a value of 0.0821 atm L K-1 mol-1

Then, apply ideal gas equation

pV = nRT

1.003 atm x V = 3.00 moles x 0.0821 atm L K-1 mol-1 x 297K

1.003 atm•V = 73.15 atm•L

Divide both sides by 1.003 atm

1.003 atm•V/1.003 atm = 73.15 atm•L/1.003 atm

V = 72.93 L

Thus, the volume of the gas is 72.93 litres

5 0
3 years ago
Using the phase diagram for H2O, which of the following correctly describes water at 0°C and 1 atm?
SCORPION-xisa [38]

At the melting point. Draw a line up from 0 degrees and a line to the right from 1 atm. They meet at the line between solid and liquid... the melting point

6 0
3 years ago
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g A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is __________ compared to a 300 mOsM sol
ICE Princess25 [194]

Answer:

A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is hypotonic compared to a 300 mOsM solution and hypotonic compared to a cell with 300 mOsM (non-penetrating solutes) interior.

Explanation:

The osmolarity is calculated from the molar concentration of the active particles in the solution. We have a solution that is composed of NaCl, CaCl₂ and urea.

When they are dissolved in water, they dissociate into particles as follows:

NaCl → Na⁺ + Cl⁻  (2 particles per compound)

CaCl₂ → Ca²⁺ + 2 Cl⁻ (3 particles per compound)

urea: not dissociation (1 particle per compound)

Then, we have to calculate the osmolarity of the solution. We multiply the molarity of each compound by the number of particles produced by the compound in water:

Osm = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) + (20 mM urea x 1) = 280 mOsm

Compared with 300 mOsm, 280 mOsm has a lower osmolarity, so it is a hypotonic solution.

To compare with a cell's osmolarity, we have to consider only the non-penetrating solutes. Urea is considered a penetrating solute for mammalian cells. So, the osmolarity of non-penetrating solutes (NaCl  and CaCl₂) is calculated as:

Osm (non-penetrating solutes) = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) = 260 mOsm

Therefore, we have:

Compared to 300 mOsm solution ⇒ 280 mOsm solution is a hypotonic solution

Compared to a cell with 300 mOsm ⇒ 260 mOsm solution is hypotonic

4 0
2 years ago
What coefficient values will balance the reaction shown? CS2 + Cl2? CCl4 + S2Cl2 A. 1,1,1,1 B. 1,3,1,1 C. 2,1,1,1 D. 1,2,1,2
Talja [164]
The answer is B. 1,3,1,1.

Brady
5 0
3 years ago
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