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3 years ago

What could you change or add to your models based on what you have learned about molecules and phases so far?

2 answers:

5 0

Answer:Phase changes require either the addition of heat energy (melting, evaporation, and sublimation) or subtraction of heat energy (condensation and freezing). ... Changing the amount of heat energy usually causes a temperature change.

Explanation:

babymother [125]3 years ago

3 0

Ummm I don’t know ‍♀️

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A solution is prepared by mixing 200.0 g of water, H2O, and 300.0 g of

VikaD [51]

Answer:

Mole Fraction (H₂O) = 0.6303

Mole Fraction (C₂H₅OH) = 0.3697

Explanation:

(Step 1)

Calculate the mole value of each substance using their molar masses.

Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol

Molar Mass (H₂O): 18.014 g/mol

200.0 g H₂O 1 mole
--------------------- x ------------------ = 11.10 moles H₂O
18.014 g

Molar Mass (C₂H₅OH): 2(12.011 g/mol) + 6(1.008 g/mol) + 15.998 g/mol

Molar Mass (C₂H₅OH): 46.068 g/mol

300.0 g C₂H₅OH 1 mole
---------------------------- x -------------------- = 6.512 moles C₂H₅OH
46.068 g

(Step 2)

Using the mole fraction ratio, calculate the mole fraction of each substance.

moles solute
Mole Fraction = ------------------------------------------------
moles solute + moles solvent

11.10 moles H₂O
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH

Mole Fraction (H₂O) = 0.6303

6.512 moles C₂H₅OH
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH

Mole Fraction (C₂H₅OH) = 0.3697

7 0

2 years ago

What is the amount in grams of EDTA needed to make 329.5 mL of a 0.03 M EDTA solution. The molar mass of EDTA is 374 g/mol. (Onl

12345 [234]

Taking into account the definition of molarity and molar mass, the amount of mass of EDTA needed to make 329.5 mL of a 0.03 M EDTA solution is 3.69699 grams.

<h3>Definition of molarity</h3>

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

$molarity=\frac{number of moles}{volume}$

Molarity is expressed in units .

Definition of molar mass

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

In this case, you need to make 329.5 mL of a 0.03 M EDTA solution. Then, you know:

Molarity= 0.03 M
number of moles= ?
Volume= 329.5 mL= 0.3295 L (being 1000 mL= 1 L)

Replacing in the definition of molarity:

$0.03 M=\frac{number of moles}{0.3295 L}$

Solving:

0.03 M× 0.3295 L= number of moles

number of moles= 0.009885 moles

Now, knowing that the molar mass of EDTA is 374 $\frac{grams}{mole}$ , the mass of EDTA needed is calculated by:

0.009885 moles×374 $\frac{grams}{mole}$ = 3.69699 grams

Finally, the amount of mass of EDTA needed to make 329.5 mL of a 0.03 M EDTA solution is 3.69699 grams.

Learn more about

molarity:

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molar mass:

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5 0

2 years ago

Why are mercury thermometers rarely used in the laboratory?

ozzi

Because it is one of the strongest chemicals and it is very dangerous.

8 0

4 years ago

Read 2 more answers

5. The compound is called:

motikmotik

Answer:

Vanadium (VI) Phosphide

Explanation:

Vanadium has a 6+ charge for this compound, so there needs to be two phosphorus atoms to satisfy the charges. This is an ionic compound as well so the roman numerals need to be present.

4 0

3 years ago

Calculate the pH of each solution.

WITCHER [35]

Answer:

See Explanations

Explanation:

pH =-log[H₃O⁺] = -log[H⁺]

pOH = -log[OH⁻]

For weak acids [H⁺] = SqrRt(Ka·[Acid])

For weak bases [OH⁻] = SqrRt(Kb·[Base])

pH + pOH = 14

__________________________________________

A. Given 0.18M CH₃NH₂; Kb = (4.4 x 10⁻⁴)* => pH = 11.95

CH₃NH₂ + H₂O => CH₃NH₃OH ⇄ CH₃NH₃⁺ + OH⁻;

[OH⁻] = SqrRt(Kb·[weak base]) = SqrRt(4.4 x 10⁻⁴ x 0.18)M = 8.97 x 10⁻³M

=> pOH = -log[OH⁻] = -log(8.93x10⁻³) = -(-2.05) = 2.05

=> pH = 14 - pOH = 14 - 2.05 = 11.95.

*Kb values for most ammonia derivatives in water can be found online by searching 'Kb-values for weak bases'. Kb-values for methyl amine and methylammonium chloride are both 4.4x10⁻⁴.

___________________________________________________

B. Given 0.18M CH₃NH₃Cl

In water ... CH₃NH₃Cl => CH₃NH₃⁺ + Cl⁻; Kb(CH₃NH₃Cl) = 4.4 x 10⁻⁴

Cl⁻ + H₂O => No Rxn (i.e.; no hydrolysis occurs) ... Cl⁻ does not react with H₂O.

Hydrolysis Reaction of Methylammonium Ion:

CH₃NH₃⁺ + H₂O => CH₃NH₄OH ⇄ CH₃NH₄⁺ + OH⁻

Ka' x Kb = Kw => Ka' = Kw/Kb = 10⁻¹⁴/4.4 x 10⁻⁴ = 2.27 x 10⁻¹¹ Ka' = [CH₃NH₄⁺][OH⁻]/[CH₃NH₄OH] = (x)(x)/(0.18M) = (x²/0.18M) = 2.27 x 10⁻¹¹ => x = [OH⁻] = SqrRt(2.27x10⁻¹¹ x 0.18)M = 2.02 x 10⁻⁶M => pOH = -log(2.02 x 10⁻⁶) = -(-5.69) = 5.69 => pH = 14 - pOH = 14 - 5.69 = 8.31.

*note => the general nature of halide interactions would increase acidity (lower pH) of the halogenated compound.

C. A mixture of 0.18M CH₃NH₂ and 0.18M CH₃NH₃Cl

Mixture of 0.18M CH₃NH₂ + 0.18M CH₃NH₃Cl

In Water ...

=> 0.18M CH₃NH₃OH + 0.18M CH₃NH₃Cl

=> 0.18M CH₃NH₃⁺ + 0.1M OH⁻ + 0.18M CH₃NH₃⁺ + 0.18M Cl⁻

=> 0.36M CH₃NH₃⁺ + 0.18M OH⁻ + 0.18M Cl⁻

-----------------------------------------------------------

Ka'(CH₃NH₃⁺) x Kb(CH₃NH₂) = Kw => Ka'(CH₃NH₃⁺) = Kw/Kb(CH₃NH₂)

=> Ka'(CH₃NH₃⁺) = (10⁻¹⁴/4.4x10⁻⁴) = 2.27x10⁻¹¹

----------------------------------------------------------

From the 0.36M CH₃NH₃⁺

=> CH₃NH₃⁺ + H₂O ⇄ CH₃NH₄⁺ + OH⁻

C(eq) 0.36M ---- x x (<= at equilibrium after mixing)

Ka'(CH₃NH₃⁺) = [CH₃NH₄⁺][OH⁻]/[CH₃NH₃⁺] = x²/(0.36M)

=> x = [OH⁻] = SqrRt(Ka'(CH₃NH₃⁺)·0.36M) = SqrRt(2.27x10⁻¹¹/0.36) = 0.0126M

=> Total [OH⁻] = 0.0126M + 0.18M = 0.1926M from hydrolysis process

=> final solution mix is therefore, 0.1926M in OH⁻ + 0.18M in Cl⁻

--------------------------------------------------------

Cl⁻ + H₂O => No Rxn (Cl⁻ does not react with H₂O)

The 0.1926M in OH⁻ => [H⁺] = Kw/[OH⁻] = (10⁻¹⁴/0.1926)M = 5.192 x 10⁻¹⁴M in H₃O⁺ ions (= H⁺ ions) ...

∴pH = -log[H⁺] = -log(5.192x10⁻¹⁴) = -(-13.29) = 13.29 for solution mix

6 0

3 years ago