Determine the energy in joules of a photon whose frequency is 3.55 x10^17 hz( with units)

Zinc Sulfide reacts with oxygen according to the reaction:

zhuklara [117]

The amount in moles of the excess reactant left is 0.5 mole

<h3>Balanced equation</h3>

2ZnS (s) + 3O₂(g) --> 2ZnO (s) + 2SO₂(g)

From the balanced equation,

2 moles of ZnS reacted with 3 moles of O₂

<h3>How to determine the excess reactant</h3>

From the balanced equation,

2 moles of ZnS reacted with 3 moles of O₂

Therefore,

4.2 moles of ZnS will react with =(4.2 × 3) / 2 = 6.3 moles of O₂

From the calculations made above, we can see that only 6.3 moles of O₂ out of 6.8 moles given, is required to react completely with 4.2 moles of ZnS.

Thus, ZnS is the limiting reactant and O₂ is the excess reactant.

<h3> How to determine the mole of the excess reactant remaining</h3>

The excess reactant is O₂. Thus the mole remaining after the reaction can be obtained as illustrated below:

Mole of O₂ given = 6.8 moles
Mole of O₂ that reacted = 6.3 moles
Mole of O₂ remaining =?

Mole of O₂ remaining = (Mole of O₂ given) - (Mole of O₂ that reacted)

Mole of O₂ remaining = 6.8 - 6.3

Mole of O₂ remaining = 0.5 mole

Learn more about stoichiometry:

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6 0

1 year ago

Which of these actions would increase heat transfer between two objects?

kicyunya [14]

I believe it would be the 2nd one, “increasing the area of their contact”.

8 0

3 years ago

Read 2 more answers

What weight of sodium formate must be added to 4.00L of 1.00 M formic acid to produce a buffer solution that has a pH of 3.50?

tester [92]

Answer:

156,4 g of sodium formate

Explanation:

The pka of the formic acid is 3,74. Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [A⁻] / [HA] <em>(1)</em>

Where A⁻ is the conjugate base (Formate) and HA is the formic acid.

4.00L of 1.00M formic acid contain:

4.00L × (1.00mol /L) = 4.00 moles

Replacing these moles, the desired pH and the pka value in (1):

3,50 = 3,74 log₁₀ [A⁻] / 4,00 moles

-0,24 = log₁₀ [A⁻] / 4,00 moles

0,575 = [A⁻] / 4,00 moles

<em>2,30 moles = [A⁻] </em>

That means you need 2,30 moles of formate (Sodium formate), to produce the desired buffer. As the molar mass of sodium formate is 68,01g/mol, the weight of sodium formate that must be added is:

2,30mol×(68,01g/mol) =<em> 156,4 g of sodium formate</em>.

I hope it helps!

7 0

2 years ago

Calculate the mass percentage of Na2SO4 in a solution containing 10.9 g Na2SO4 in 482 g water.

boyakko [2]

Answer:

The answer to your question is 2.21 %

Explanation:

Data

mass of Na₂SO₄ = 10.9 g

mass of H₂O = 482 g

Formula

Percent by mass = $\frac{mass of solute}{total mass}$ x 100

Substitution

$Percent by mass = \frac{10.9}{10.9 + 482}$ x 100

Simplification and result

$Percent by mass = \frac{10.9}{492.9}$ x 100

Percent by mass = 0.022 x 100

Percent by mass = 2.21

3 0

3 years ago

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Compute the equilibrium constant at 25 ∘C for the reaction between Sn2+(aq) and Cd(s), which form Sn(s) and Cd2+(aq). Express yo

Radda [10]

Answer:

6.1×10^8

Explanation:

The reaction is;

Sn^2+(aq) + Cd(s) -----> Sn(s) + Cd^2+(aq)

E°cell = E°cathode - E°anode

E°cathode= -0.14 V

E°anode= -0.40 V

E°cell = -0.14-(-0.40)

E°cell= -0.14+0.40

E°cell= 0.26 V

But

E°cell= 0.0592/n log K

E°cell= 0.0592/2 log K

0.26= 0.0296log K

log K = 0.26/0.0296

log K= 8.7838

K= Antilog (8.7838)

K= 6.1×10^8

8 0

3 years ago