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3 years ago

Which of the following are likely to form an ionic bond

Chemistry

1 answer:

sattari [20]3 years ago

3 0

Answer:

No question is posted.

But Ionic bond is mostly between metals live Na, K, Ca, Mg and strong non metals like Cl, O, F

Explanation:

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Why would methane attain a tetrahedral structure?

lys-0071 [83]

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3 0

3 years ago

Read 2 more answers

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

3 years ago

Please help me with this thank you!

ludmilkaskok [199]

Answer:

1. B

2. C

3. A

4. E

5. D

6. A, B

7. C, E

8. D

5 0

2 years ago

Mole-Mass Conversions

Vanyuwa [196]

1.) 28.0 grams of oxygen
28 grams (1 mole/16 grams per mole)=1.75 moles oxygen
2.)5.0 moles of Iron
5 moles(55.845 grams/1 mole)=279.225 grams Iron
3.) 452 grams Argon
452 grams(1 mole/39.948 grams)=11.315 moles Argon
4.) 16.5 moles Hydrogen
16.5 moles(1.01 grams/1 mole)=16.665 grams Hydrogen

3 0

3 years ago

g Suppose 0.0350 g M g is reacted with 10.00 mL of 6 M H C l to produce aqueous magnesium chloride and hydrogen gas. M g ( s ) +

iren2701 [21]

Answer:

Mg will be the limiting reagent.

Explanation:

The balanced reaction is:

Mg + 2 HCl → MgCl₂ + H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Mg: 1 mole
HCl: 2 moles
MgCl₂: 1 mole
H₂: 1 mole

Being the molar mass of each compound:

Mg: 24.3 g/mole
HCl: 36.45 g/mole
MgCl₂: 95.2 g/mole
H₂: 2 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Mg: 1 mole* 24.3 g/mole= 24.3 g
HCl: 2 moles* 36.45 g/mole= 72.9 g
MgCl₂: 1 mole* 95.2 g/mole= 95.2 g
H₂: 1 mole* 2 g/mole= 2 g

0.0350 g of Mg is reacted with 10.00 mL (equal to 0.01 L) of 6 M HCl.

Molarity being the number of moles of solute that are dissolved in a certain volume, expressed as:

$Molarity=\frac{number of moles of solute}{volume}$

in units $\frac{moles}{liter}$

then, the number of moles of HCl that react is:

$6 M=\frac{number of moles of HCl}{0.01 L}$

number of moles of HCl= 6 M*0.01 L

number of moles of HCl= 0.06 moles

Then you can apply the following rule of three: if by stoichiometry 2 moles of HCl react with 24.3 grams of Mg, 0.06 moles of HCl react with how much mass of Mg?

$mass of Mg=\frac{0.06 moles of HCl* 24.3 grams of Mg}{2 moles of HCl}$

mass of Mg= 0.729 grams

But 0.729 grams of Mg are not available, 0.0350 grams are available. Since you have less mass than you need to react with 0.06 moles of HCl, <u><em>Mg will be the limiting reagent.</em></u>

7 0

3 years ago