3 years ago

A 4.00-gram sample of sodium chloride was added to a beaker containing 40.0 mL 0.033 M lead(II)

Chemistry

1 answer:

melamori03 [73]3 years ago

6 0

The answer is going to be 2.5

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When a substance changes from one phase to another, which of the following occurs?

Shkiper50 [21]

Answer:

Explanation:

The average kinetic energy of the Substance changes.

8 0

2 years ago

Calculate the pH of a solution created by placing 2.0 grams of yttrium hydroxide, Y(OH)3, in 2.0 L of H2O. Ksp for Y(OH)3 is 6.0

oee [108]

Answer:

pH = 8.314

Explanation:

Y(OH)3(s) ↔ Y+ + 3OH-

equil: S S 3S

∴ Ksp = [ Y+ ] * [ OH- ]³ = 6.0 E-24

⇒ 6.0 E-24 = ( S )*( 3S )³

⇒ 6.0 E-24 = 27S∧4

⇒ 2.22 E-25 = S∧4

⇒ ( 2.22 E-25 )∧(1/4) = S

⇒ S = 6.866 E-7 M

⇒ [ OH- ] = 3*S =2.06 E-6 M

⇒ pOH = - Log [ OH- ]

⇒ pOH = - Log ( 2.06 E-6 )

⇒ pOH = 5.686

∴ pH = 14 - pOH

⇒ pH = 8.314

8 0

3 years ago

What do you think is happening at the nanoscale that keeps the salad dressing with an emulsifier mixed?

saveliy_v [14]

Answer:

✨ science ✨ and ✨ big brain stuff ✨

Explanation:

sorry

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2 years ago

In a sample of solid ba(no3)2 the ratio of barium ions to nitrate ions is

Nata [24]

I believe the ratio is 1:2

7 0

3 years ago

Just as one dozen eggs always has 12 eggs in it, one mole of a

zzz [600]

Answer:

$6.022x10^{23}atoms \ Al$

Explanation:

Hello,

In this case, given the described concept regarding the Avogadro's number, we can easily notice that 27.0 g of aluminium foil has 6.022x10²³ atoms as shown below based on the mass-mole-particles relationship:

$27.0gAl*\frac{1molAl}{27.0gAl} *\frac{6.022x10^{23}atoms \ Al}{1molAl} \\\\=6.022x10^{23}atoms \ Al$

Notice this is backed up by the fact that aluminium molar mass if 27.0 g/mol.

Best regards.

8 0

3 years ago