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evablogger [386]
3 years ago
14

Energy from the Sun reaches Earth through

Chemistry
2 answers:
g100num [7]3 years ago
8 0

Solar Radiation(Basically Radiation)

zheka24 [161]3 years ago
3 0
Solar radiation is the answer
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If I add water to oil what will happen
Alex

The oil, even a little drop, will spread out over the water surface and break up into many little blobs.

8 0
3 years ago
The standard reduction potentials for the Ag+|Ag(s) and Zn2+| Zn(s) half-cell reactions are +0.799 V and -0.762 V, respectively.
mihalych1998 [28]

<u>Answer:</u> The potential of the given cell is 1.551 V

<u>Explanation:</u>

The given chemical cell follows:

Zn(s)|Zn^{2+}(0.125M)||Ag^{+}(0.240M)|Ag(s)

<u>Oxidation half reaction:</u> Zn(s)\rightarrow Zn^{2+}(0.125M)+2e^-;E^o_{Zn^{2+}/Zn}=-0.762V

<u>Reduction half reaction:</u> Ag^{+}(0.240M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.799V       ( × 2)

<u>Net cell reaction:</u> Zn(s)+2Ag^{+}(0.240M)\rightarrow Zn^{2+}(0.125M)+2Ag(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.799-(-0.762)=1.561V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Ag^{+}]^2}

where,

E_{cell} = electrode potential of the cell = ? V

E^o_{cell} = standard electrode potential of the cell = +1.561 V

n = number of electrons exchanged = 2

[Zn^{2+}]=0.125M

[Ag^{+}]=0.240M

Putting values in above equation, we get:

E_{cell}=1.561-\frac{0.059}{2}\times \log(\frac{(0.125)}{(0.240)^2})

E_{cell}=1.551V

Hence, the potential of the given cell is 1.551 V

6 0
4 years ago
Calculate the following:
d1i1m1o1n [39]

Answer:

Explanation:

1) Given data:

Number of moles of lead = 4.3×10⁻³ mol

Mass of lead = ?

Solution:

Mass = number of moles × molar mass

Molar mass of lead = 207.2 g/mol

Mass = 4.3×10⁻³ mol × 207.2 g/mol

Mass =  890.96 g

2) Given data:

Number of atoms of antimony = 3.8×10²² atoms

Mass of antimony = ?

Solution:

1 mole contain 6.022 ×10²³ atoms

3.8×10²² atoms × 1 mol / 6.022 ×10²³ atoms

0.63×10⁻¹ mol

0.063 mol

Mass = number of moles × molar mass

Molar mass of lead = 121.76 g/mol

Mass = 0.063 mol ×  121.76 g/mol

Mass =  7.67 g

3) Given data:

Mass of tungsten = 15.5 Kg (15.5 kg × 1000 g/ 1kg = 15500 g)

Number of atoms = ?

Solution:

Number of moles of tungsten:

Number of moles = mass/molar mass

Number of moles = 15500 g / 183.84 g/mol

Number of moles = 84.3 mol

1 mole contain 6.022 ×10²³ atoms

84.3 mol  × 6.022 ×10²³ atoms / 1mol

507.65 ×10²³ atoms

8 0
3 years ago
Can our bodies produce salt also is it possible to use our body oil to cook
GaryK [48]
No because your body can't take that much salt
3 0
3 years ago
I don’t wanna fail if you know the answer pls help me :(
stealth61 [152]

Answer:

liquid and gas........................

7 0
3 years ago
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