3x10^8 / (670.8 * 10^-9) =4.47x10^14 Hz 4.47x10^14 Hz multiplied by plank's constant = 2.9634x10^-19
2.96 x10-19 J
Answer:
78.04g of 0.448 moles must be added
Explanation:
Using H-H equation we can find the pH of the buffer:
pH = pKa + log [A⁻] / [HA]
<em>Where pH is the pH of the buffer = 7.2</em>
<em>pKa = 7.1</em>
<em>[A⁻] = [K₂HPO₄]</em>
<em>[HA] = [KH₂PO₄]</em>
<em />
Replacing:
7.2 = 7.1 + log [K₂HPO₄] / [KH₂PO₄]
0.1 = log [K₂HPO₄] / [KH₂PO₄]
1.2589 = [K₂HPO₄] / [KH₂PO₄] <em>(1)</em>
<em />
And as the concentration of the buffer is:
1M = [K₂HPO₄] + [KH₂PO₄] <em>(2)</em>
<em></em>
Replacing (2) in (1):
1.2589 = 1M - [KH₂PO₄] / [KH₂PO₄]
1.2589 [KH₂PO₄] = 1M - [KH₂PO₄]
2.2589 [KH₂PO₄] = 1M
[KH₂PO₄] = 0.44M
And [K₂HPO₄] = 0.56M
In 800mL = 0.8L:
0.8L * (0.56mol / L) = 0.448 moles K₂HPO₄. The mass is -Molar mass K₂HPO₄: 174.2g/mol-:
0.448 moles * (174.2g / mol) =
<h3>78.04g of 0.448 moles must be added</h3>
Answer:
25.9 g
Explanation:
Given that,
No of moles of calcium phosphate,
We need to find how many grams of calcium phosphate has this much of no of moles.
Mass divided by molar mass is equal to the no of moles on a molecule.
The molar mass of calcium phosphate is 310.18 g/mol
Using the concept of no of moles as follows :
Out of given options, option (a) i.e. 25.9 g is the correct answer.
A.
Cr⁺¹ + Sn⁺⁴ ⇒ Cr⁺³ + Sn⁺²
Cr⁺¹ ⇒ Cr⁺³ + 2e⁻
The half-reaction of oxidation: chrome is an electron donor, losses 2 electrons ie it is oxidized
2e⁻ + Sn⁺⁴ ⇒ Sn⁺²
The half-reaction of reduction: tin is an electron acceptor, receives 2 electrons, ie it is reduced
b.
3Hg⁺² + 2Fe ⇒ 3Hg + 2Fe⁺³
2Fe ⇒ 2Fe⁺³ + 6e⁻
The half-reaction of oxidation: iron is an electron donor, losses 3 electrons ie it is oxidized
6e⁻ + 3Hg⁺² + ⇒ 3Hg
The half-reaction of reduction: mercury is an electron acceptor, receives 2 electrons, ie it is reduced
<span>
c.
2As + 3Cl</span>₂ ⇒ 2AsCl₃
2As ⇒ 2As⁺³ + 6e⁻
The half-reaction of oxidation: arsenic is an electron donor, losses 3 electrons ie it is oxidized
6e⁻ + 3Cl₂ ⇒ 6Cl⁻
The half-reaction of reduction: chlorine is an electron acceptor, receives 1 electron, ie it is reduced
<span>
d.
NaBr + Cl</span>₂ ⇒ NaCl + Br₂<span>
</span>2Br⁻ ⇒ Br₂ + 2e⁻
The half-reaction of oxidation: bromine is an electron donor, losses 1 electron ie it is oxidized
2e⁻ + Cl₂ ⇒ 2Cl⁻
The half-reaction of reduction: chlorine is an electron acceptor, receives 1 electron, ie it is reduced
e.
Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂
3C⁺² ⇒ 3C⁺⁴ + 6e⁻
The half-reaction of oxidation: carbon is an electron donor, losses 2 electrons ie it is oxidized
6e⁻ + 2Fe⁺³ ⇒ 2Fe
The half-reaction of reduction: iron is an electron acceptor, receives 3 electrons, ie it is reduced
Answer:
7 moles of water will produced from 3.5 moles of ammonium nitrate.
Explanation:
Given data:
Moles of ammonium nitrate = 3.5 mol
Moles of water produced = ?
Solution:
Chemical equation:
2NH₄NO₃ → 2N₂ + 4H₂O + O₂
Now we will compare the moles of ammonium nitrate with water.
NH₄NO₃ : H₂O
2 : 4
3.5 : 4/2×3.5 = 7 moles
Thus 7 moles of water will produced from 3.5 moles of ammonium nitrate.