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matrenka [14]
3 years ago
13

What is NADPH2 ? where is it used?

Chemistry
1 answer:
uysha [10]3 years ago
3 0
<span>NADPH2 (or NADPH) - nicotinamide adenine dinucleotide phosphate</span>

NADPH2 transfers hydrogen in the process of photosynthesis
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After substantial heating, 6.25 g of iron produced 18.00 g of a compound with chlorine. The empirical formula is:
babunello [35]

Answer:

Option A. FeCl3

Explanation:

The following data were obtained from the question:

Mass of iron (Fe) = 6.25g

Mass of the compound formed = 18g

From the question, we were told that the compound formed contains chlorine. Therefore the mass of chlorine is obtained as follow

Mass of chlorine (Cl) = Mass of compound formed – Mass of iron.

Mass of chlorine (Cl) = 18 – 6.25

Mass of chlorine (Cl) = 11.75g

The compound therefore contains:

Iron (Fe) = 6.25g

Chlorine (Cl) = 11.75g

The empirical formula for the compound can be obtained by doing the following:

Step 1:

Divide by their molar mass

Fe = 6.25/56 = 0.112

Cl = 11.75/35.5 = 0.331

Step 2:

Divide by the smallest

Fe = 0.112/0.112 = 1

Cl = 0.331/0.112 = 3

The empirical formula for the compound is FeCl3

3 0
3 years ago
"M" represents a metallic element, the oxide of which has the formula M2O. The formula of the chloride of M is..................
Rus_ich [418]
The formula of the chloride of M will be MCI2.
8 0
3 years ago
A galvanic cell consists of one half-cell that contains Ag(s) and Ag+(aq), and one half-cell that contains Cu(s) and Cu2+(aq). W
Agata [3.3K]

Answer : The 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Explanation :

Galvanic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the voltaic cell or electrochemical cell.

In the galvanic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

We are taking the value of standard reduction potential form the standard table.

E^0_{[Ag^{+}/Ag]}=+0.80V

E^0_{[Cu^{2+}/Cu]}=+0.34V

In this cell, the component that has lower standard reduction potential gets oxidized and that is added to the anode electrode. The second forms the cathode electrode.

The balanced two-half reactions will be,

Oxidation half reaction (Anode) : Cu(s)\rightarrow Cu^{2+}(aq)+2e^-

Reduction half reaction (Cathode) : Ag^{+}(aq)+e^-\rightarrow Ag(s)

Thus the overall reaction will be,

Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

From this we conclude that, 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Hence, the 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

7 0
3 years ago
The equation most often associated with Newton's Second Law of Motion is
nignag [31]

Answer:

F = ma or Force = mass x acceleration.

Explanation:

Newtons second law is about the force put onto an object. To find this you would use the formula:

Force is equal to mass times acceleration (F=ma)

7 0
3 years ago
Read 2 more answers
0.53g of acetanilide was subjected to kjeldahl determination and the ammonia produced was collected in 50cm3 of 0.50M of h2so4.o
Lady bird [3.3K]

Answer:

10.57% of N in acetanilide

Explanation:

All nitrogen in the sample is converted in NH₃ in the Kjeldahl determination. The NH₃ reacts with H₂SO₄ as follows:

2NH₃ + H₂SO₄ → 2NH₄⁺ + SO₄²⁻

The acid in excess in titrated with Na₂CO₃ as follows:

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

To solve this question we must find the moles of sodium carbonate = Moles of H₂SO₄ in excess. The added moles - Moles in excess = Moles of sulfuric acid that reacts:

<em>Moles Na₂CO₃ anf Moles H₂SO₄ in excess:</em>

0.025L * (0.05mol / L) = 1.25x10⁻³ moles Na₂CO₃ / 0.01360L =

0.09191M * 0.250L = 0.0230 moles H₂SO₄ in excess.

<em>Moles H₂SO₄ added:</em>

0.050L * (0.50mol / L) = 0.0250 moles H₂SO₄ added

<em>Moles that react:</em>

0.0250 moles - 0.0230 moles = 0.0020 moles H₂SO₄

<em>Moles of NH₃ = Moles N:</em>

0.0020 moles H₂SO₄ * (2mol NH₃ / 1mol H₂SO₄) = 0.0040 moles NH₃ = Moles N

<em>mass N and mass percent:</em>

0.0040 moles N * (14g / mol) = 0.056gN / 0.53g * 100 =

<h3>10.57% of N in acetanilide</h3>
7 0
2 years ago
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