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3 years ago

Calculate the molarity of each solution.

1 answer:

8 0

Q1)
molarity is defined as the number of moles of solute in 1 L solution
the number of moles of LiNO₃ - 0.38 mol
volume of solution - 6.14 L
since molarity is number of moles in 1 L
number of moles in 6.14 L - 0.38 mol
therefore number of moles in 1 L - 0.38 mol / 6.14 L = 0.0619 mol/L
molarity of solution is 0.0619 M

Q2)
the mass of C₂H₆O in the solution is 72.8 g
molar mass of C₂H₆O is 46 g/mol
number of moles = mass present / molar mass of compound
the number of moles of C₂H₆O - 72.8 g / 46 g/mol
number of C₂H₆O moles - 1.58 mol
volume of solution - 2.34 L
number of moles in 2.34 L - 1.58 mol
therefore number of moles in 1 L - 1.58 mol / 2.34 L = 0.675 M
molarity of C₂H₆O is 0.675 M

Q3)

Mass of KI in solution - 12.87 x 10⁻³ g
molar mass - 166 g/mol
number of mole of KI = mass present / molar mass of KI
number of KI moles = 12.87 x 10⁻³ g / 166 g/mol = 0.0775 x 10⁻³ mol
volume of solution - 112.4 mL
number of moles of KI in 112.4 mL - 0.0775 x 10⁻³ mol
therefore number of moles in 1000 mL- 0.0775 x 10⁻³ mol / 112.4 mL x 1000 mL
molarity of KI - 6.90 x 10⁻⁴ M

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How many grams of product are formed from 2.0 mol of N2 (g) and 8.0 mol of Mg(s)? Show all calculations leading to an answer. Li

Vaselesa [24]

Balanced chemical reaction happening here is:

3Mg(s) + N₂(g) → Mg₃N₂(s)

moles of product formed from each reactant:

2.0 mol of N2 (g) x 1 mol Mg₃N₂ = 2 mol Mg₃N₂

1 mol N2

and

8.0 mol of Mg(s) x 1 mol Mg₃N₂ = 2.67 mol Mg₃N₂

3 mol Mg

Since N2 is giving the least amount of product(Mg₃N₂) ie. 2 mol Mg₃N₂

N2 is the limiting reactant here and Mg is excess reactant.

Hence mole of product formed here is 2 mol Mg₃N₂

molar mass of Mg₃N₂

= 3 Mg + 2 N

= 101g/mol

mass of product(Mg₃N₂) formed

= moles x Molar mass

= 2 x 101

= 202g Mg₃N₂

202g of product are formed from 2.0 mol of N2(g) and 8.0 mol of Mg(s).

The following are indicators of chemical changes:

Change in Temperature

Change in Color

Formation of a Precipitate

8 0

3 years ago

Avanti works in a bookstore. She has four books and is going to place them in two stacks. The diagram above shows the books befo

Feliz [49]

Answer:

Well they didn't transfer any energy when they weren’t touching and it did t produce any energy if it didn’t move. Since they are on top of each other they are causing momentum on each other creating kinetic energy

Explanation:

3 0

3 years ago

Nail polish remover containing acetone was spilled in a room 7.2 m × 5.8 m × 3.9 m. Measurements indicated that 6,400 mg of acet

sergeinik [125]

Volume of room = $7.2 m \times 5.8 m \times 3.9 m$

= $162.864 m^{3}$

Now, according to conversion factor, convert 6,400 mg to micrograms

Since, 1 mg is equal to 1000 microgram.

Therefore, 6,400 mg = $6,400 \times 1000$

= 6,400,000 micrograms of acetone.

To calculate concentration of acetone, divide volume and evaporated amount of acetone in micrograms.

Thus,

Concentration of acetone = $\frac{6,400,000 microgram}{162.864 m^{3}}$

= 39296.5910 microgram per cubic meter or $3.92\times 10^{4} \mu g/m^{3}\simeq 4.0\times 10^{4} \mu g/m^{3}$

Hence, concentration of acetone is equal to $4.0\times 10^{4} \mu g/m^{3}$

7 0

3 years ago

How does the formation of covalent bonds relate to the octet rule?

Triss [41]

Answer:

Covalent bonding occurs when atoms share valence electrons to form a chemical bond. Atoms of nonmetals form covalent bonds by sharing electrons in order to achieve an octet of valence electrons, like the noble gases, except for helium, which only has two valence electrons. The octet refers to the highest energy s and p sublevels, the valence shell, which in noble gases, except for helium, are filled with eight electrons, two in the s sublevel and six in the p sublevel.

7 0

3 years ago

Help!!!

kupik [55]

Answer:

1. 0.82 gram of Ag+

2. 4.79 g of Ag₂O₃S

Explanation:

From the given information:

Total amount of Ag₂O₃S = 2.24 grams

Atomic mass of Ag+ =107.86 g/mole

molar mass of Ag₂O₃S = 295.8 g/mole

∴

The mass of the Silver (Ag) in grams is:

$= Total\ amount \ of \ Ag^+ \times \dfrac{107.86 \ g/mol}{295.8 \ g/mol}$

$=2.24 \times \dfrac{107.86 \ g/mol}{295.8 \ g/mol}$

= 0.82 gram of Ag+

Here, the total amount of Ag₂O₃S = unknown

Atomic mass of Ag+ = 107.86 g/mole

molar mass of Ag₂O₃S = 295.8 g/mole

amount of Ag+ = 1.75 g

∴

The mass of Ag₂O₃S = $Total \ amount \ of \ Ag^+ \times \dfrac{295.8 \ g/mol}{107.86\ g/mol}$

$=1.75 \times \dfrac{295.8 \ g/mol}{107.86\ g/mol}$

= 4.79 g of Ag₂O₃S

4 0

3 years ago