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yawa3891 [41]
3 years ago
11

How many grams of CaCl, are in 250 mL of 2.0 M CaCl ?

Chemistry
1 answer:
stellarik [79]3 years ago
6 0

Answer: 55 g CaCl2 hope this helps

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When a gas is given off as a liquid boils it is an example of a?
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This is an example of a condenser
7 0
3 years ago
Given that nitrogen forms three bonds with hydrogen to make <img src="https://tex.z-dn.net/?f=NH_%7B3%7D" id="TexFormula1" title
lbvjy [14]

Answer:

Three hydrogen atoms to form PH₃.

Explanation:

Hello!

In this case, since the elements belonging to the nitrogen family (N, P, As, Sb and Bi) show five valence electrons, because there are five electrons at their outer shell, it is clear that if phosphorous bonds with hydrogen, it is going to require the same amount of oxygen atoms (3) because elements having five valence electrons need 3 bonds in order to attain the octet (5+3=8).

Therefore the compound would be:

PH_3

Which is phosphine.

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3 0
2 years ago
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

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