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Vinvika [58]
3 years ago
5

During photosynthesis, plants use carbon dioxide, water, and light energy to produce glucose, C

Chemistry
1 answer:
Akimi4 [234]3 years ago
3 0
The outcome of the equation shows that there is no light energy proving that light energy was absorbed to get CH20; +602
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Question 11 (1 point)
iren2701 [21]
Blue to red is acidic, red to blue is basic and no change is neutral.
6 0
3 years ago
How many Electrons, Neutrons & Protons do
EastWind [94]

Answer:

sodium

protons:11

neutrons:12

electrons:11

nitrogen

protons:7

neutrons:7

electrons:7

6 0
3 years ago
Read 2 more answers
What is the percent, by mass, of water in MgSO4.2H20
iris [78.8K]

Answer:51.1%

Explanation:

Mass percent : It is defined as the mass of the given component present in the total mass of the compound. Formula used : First we have to calculate the mass of  and . Mass of  = 18 g/mole Mass of  = 7 × 18 g/mole = 126 g/mole Mass of  = 246.47 g/mole Now put all the given values in the above formula, we get the mass percent of  in . Therefore, the mass percent of  in  is, 51.1%

8 0
2 years ago
If hydrochloric acid has a [H+] of 1.2 x 10-2 M, what is the pH?<br> 1.9<br> 0 1<br> 1.2<br> O<br> 8
Vanyuwa [196]

Answer:

<h2>1.9</h2>

Explanation:

The pH of a solution can be found by using the formula

pH = - log [ {H}^{+} ]

From the question we have

pH =  -  log(1.2 \times  {10}^{ - 2} )  \\  = 1.920818...

We have the final answer as

<h3>1.9</h3>

Hope this helps you

7 0
2 years ago
Read 2 more answers
The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate
Cloud [144]

Answer : The \Delta G for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,  

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

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : Cu\rightarrow Cu^{2+}+2e^-

Reduction : 2Ag^++2e^-\rightarrow 2Ag

Now we have to calculate the Gibbs free energy.

Formula used :

\Delta G^o=-nFE^o

where,

\Delta G^o = Gibbs free energy = ?

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = 0.46 V

Now put all the given values in this formula, we get the Gibbs free energy.

\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole

Therefore, the \Delta G for this reaction is, -88780 J/mole.

7 0
3 years ago
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