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3 years ago

How many molecules are in 2.570 moles of H2

Chemistry

2 answers:

docker41 [41]3 years ago

5 0

A lot of molecules will be in 2.570 moles of H2

ankoles [38]3 years ago

3 0

1285 mybe lh not sure

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How many of the planets have an orbital period of less than one Earth year?

GrogVix [38]

Since orbital period depends on how far you are from the sun, planets closer to the sun have a orbital period less than one earth year.

These planets are Mercury and Venus

8 0

3 years ago

Atom with three<br> full orbits<br> What is it

nekit [7.7K]

Answer:

Boron. The answer is boron.

3 0

3 years ago

Read 2 more answers

Silver sulfate (Ag2SO4) is slightly soluble in water and it partly dissolves at the equilibrium according to the following balan

natka813 [3]

Answer:

a) S = 0.0152 mol/L

b) S' = 4.734 g/L

Explanation:

Ag2SO4 ↔ 2Ag+ + SO42-

S 2S S...............in the equilibrium

Ksp = 1.4 E-5 = [ Ag+ ]² * [ SO42-]

a) molar solubility:

⇒ Ksp = ( 2S) ² * S = 1.4 E-5

⇒ 4S² * S = 1.4 E-5

⇒ S = ∛ ( 1.4 E-5 / 4 )

⇒ S = 0.0152 mol/L

b) solubility ( S' ) in grams per liter:

∴ Mw Ag2SO4 = 311.799 g/mol

⇒ S' = 0.0152 mol/L * ( 311.799 g/mol )

⇒ S' = 4.734 g/L

4 0

3 years ago

Part 1

denis-greek [22]

The moles of NaOH used in the titration would be 0.00177 moles while the molarity of the vinegar will be 0.885 M

<h3>Titration calculation</h3>

Recall that: mole = molarity x volume

In this case, the molarity of the NaOH= 0.1 M

Volume of NaOH = 17.7 - 0.0 = 17.7 mL

Mole of NaOH used = 0.1 x 17.7/1000 = 0.00177 moles.

Since NaOH and vinegar have 1:1 mole ratio, the mole of vinegar will also be 0.00177 moles.

Molarity of vinegar = mole/volume = 0.00177/0.002 = 0.885 M

More on titration calculations can be found here: brainly.com/question/9226000

7 0

2 years ago

Aluminum sulfate, known as cake alum, has a wide range of uses, from dyeing leather and cloth to purifying sewage. In aqueous so

NISA [10]

Answer:

a) The chemical reaction is given as:

$6NaOH(aq)+Al_2(SO_4)_3\rightarrow 2Al(OH)_3(s)+3Na_2SO_4(aq)$

2.571 grams of aluminum hydroxide is precipitated.

Explanation:

a) The chemical reaction is given as:

$6NaOH(aq)+Al_2(SO_4)_3\rightarrow 2Al(OH)_3(s)+3Na_2SO_4(aq)$

Moles of NaOH = n

Volume of the NaOH = 185.5 mL = 0.1855 L( 1 mL =0.001 L)

Molarity of the solution = 0.533 M

$n=0.533 M\times 0.1855 mL=0.09887 mol$

Moles of aluminum = n

15.8 g of aluminum sulfate per liter.

Volume of solution = 627 mL = 0.627 L (1 mL= 0.001 L)

Mass of aluminium sulfate in solution = 15.8 g/L × 0.627 L =9.9066 g

Moles of aluminum sulfate = $\frac{9.9066 g}{342 g/mol}=0.02897 mol$

Moles of NaOH = 0.09887 mol

According to reaction, 6 moles of NaOH reacts with 1 mole of aluminum sulfate, then 0.09887 moles of NaOH will recat with :

$\frac{1}{6}\times 0.09887 mol=0.01648 mol$

This means that sodium hydroxide moles are in limiting amount.So, amount of aluminum hydroxide will depend upon moles of sodium hydroxide.

According to reaction, 6 moles of sodium hydroxide gives 2 moles of aluminium hydroxide, then 0.09887 moles of sodium hydroxide will give :

$\frac{2}{6}\times 0.09887 mol=0.03296 mol$ of aluminum hydroxide

Mass of 0.03296 moles of aluminum hydroxide:

0.03296 mol × 78 g/mol = 2.571 g

2.571 grams of aluminum hydroxide is precipitated.

8 0

3 years ago