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2 years ago

9

Solar energy travels to the Earth in many forms, such as ultraviolet radiation, infrared radiation, and visible rays. These vari

ous forms of solar energy have what?

1 answer:

galben [10]2 years ago

4 0

Radiation. Hope this Helps

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Determine the normality of the following solutions note the species of interest is H 95 g of PO4 3- in 100mL solution

Aliun [14]

Answer : The normality of the solution is, 30.006 N

Explanation :

Normality : It is defined as the number of gram equivalent of solute present in one liter of the solution.

Mathematical expression of normality is:

$\text{Normality}=\frac{\text{Gram equivalent of solute}}{\text{Volume of solution in liter}}$

or,

$\text{Normality}=\frac{\text{Weight of solute}}{\text{Equivalent weight of solute}\times \text{Volume of solution in liter}}$

First we have to calculate the equivalent weight of solute.

Molar mass of solute $PO_4^{3-}$ = 94.97 g/mole

$\text{Equivalent weight of solute}=\frac{\text{Molar mass of solute}}{\text{charge of the ion}}=\frac{94.97}{3}=31.66g.eq$

Now we have to calculate the normality of solution.

$\text{Normality}=\frac{95g}{31.66g.eq\times 0.1L}=30.006eq/L$

Therefore, the normality of the solution is, 30.006 N

5 0

2 years ago

as a result of chemical weathering, even a hard rock such as granite can be turned into __________. a) sediment b) limestone c)

kotykmax [81]

The best ansewer would be a)sediment

3 0

2 years ago

The incomplete table below shows selected characteristics of gas laws.

Elenna [48]

Explanation :

In the given case different law related to gas is given. The attached figure shows the required solution.

Boyle's law states that the pressure is inversely proportional to the volume of the gas i.e.

$P\propto \dfrac{1}{V}$

$PV=k$

k is a constant.

Charle's law states that the volume of directly proportional to the temperature of the gas.

$V\propto T$

$V=kT$

Combined gas law is the combination of the pressure, volume and the temperature of the gas i.e.

$\dfrac{PV}{T}=k$

Hence, this is the required solution.

6 0

2 years ago

Read 2 more answers

Question 17 In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas t

mars1129 [50]

Answer:

Explanation:

N₂ + 3H₂ = 2 NH₃

1 vol 2 vol

786 liters 1572 liters

786 liters of dinitrogen will result in the production of 1572 liters of ammonia

volume of ammonia V₁ = 1572 liters

temperature T₁ = 222 + 273 = 495 K

pressure = .35 atm

We shall find this volume at NTP

volume V₂ = ?

pressure = 1 atm

temperature T₂ = 273

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

$\frac{.35\times 1572}{495} =\frac{1\times V_2}{ 273 }$

$V_2 =303.44$ liter .

mol weight of ammonia = 17

At NTP mass of 22.4 liter of ammonia will have mass of 17 gm

mass of 303.44 liter of ammonia will be equal to (303.44 x 17) / 22.4 gm

= 230.28 gm

=.23 kg / sec .

Rate of production of ammonia = .23 kg /s .

5 0

2 years ago

Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b

frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ ?

(a) $Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}$

(b) $2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}$

(c) $Al^{3+}_{(aq)}+Pb_{(s)} ->Al_{(s)}+Pb^{2+}_{(aq)}$

(d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Answer: (d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Explanation: <u>Redox</u> <u>Reaction</u> is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.

Redox reactions can be represented in shorthand form called <u>cell</u> <u>notation,</u> formed by: <em><u>left side</u></em> of the salt bridge (||), which is always the <em><u>anode</u></em>, i.e., its half-equation is as an <em><u>oxidation</u></em> and <em><u>right side</u></em>, which is always <em><u>the cathode</u></em>, i.e., its half-equation is always a <em><u>reduction</u></em>.

For the cell notation: $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$

Aluminum's half-equation is oxidation:

$Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}$

For Lead, half-equation is reduction:

$Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}$

Multiply first half-equation for 2 and second half-equation by 3:

$2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}$

$3Pb^{2+}_{(aq)}+6e^{-} -> 3Pb_{(s)}$

Adding them:

$2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

The balanced redox reaction with cell notation $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ is

$2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

6 0

2 years ago