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2 years ago

How is heat transferred through radiation?

Chemistry

1 answer:

Rama09 [41]2 years ago

5 0

Answer:

Radiation is a method of heat transfer that does not rely upon any contact between the heat source and the heated object as is the case with conduction and convection. Heat can be transmitted through empty space by thermal radiation often called infrared radiation

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We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

2 years ago

What is the composition of 'Clean dry air'<br><br> - Idk what subject its just science

Luba_88 [7]

Dry air is a mixture of nitrogen, oxygen, carbon dioxide etc.
air is a mixture of gases 78% nitrogen an 21% oxygen and other components.

7 0

2 years ago

What is the energy of a wave with wavelength of 4.2 x 10-7 m. (Hint: Calculate for frequency first.)

erik [133]

Answer:

Option B. 4.74×10¯¹⁹ J.

Explanation:

The following data were obtained from the question:

Wavelength (λ) = 4.2×10¯⁷ m

Energy (E) =.?

Next, we shall determine the frequency of the wave. This can be obtained as follow:

Wavelength (λ) = 4.2×10¯⁷ m

Velocity (v) = constant = 3×10⁸ m/s

Frequency (f) =.?

v = λf

3×10⁸ = 4.2×10¯⁷ × f

Divide both side by 4.2×10¯⁷

f = 3×10⁸ / 4.2×10¯⁷

f = 7.143×10¹⁴ s¯¹

Therefore, the frequency of the wave is 7.143×10¹⁴ s¯¹.

Finally, we shall determine the energy of the wave using the following formula

E = hf

Where

E is the energy.

h is the Planck's constant

f is the frequency

Thus, the enery of the wave can be obtained as follow:

Frequency (f) = 7.143×10¹⁴ s¯¹.

Planck's constant = 6.63×10¯³⁴ Js

Energy (E) =..?

E = hf

E = 6.63×10¯³⁴ × 7.14×10¹⁴

E = 4.74×10¯¹⁹ J

Therefore, the energy of the wave is 4.74×10¯¹⁹ J.

5 0

2 years ago

What are non examples of tissue? Help please

antiseptic1488 [7]

Epithelial

I know just one

7 0

2 years ago

The ka of hf is 6.8 x 10-4. what is the ph of a 0.35 m solution of hf?

damaskus [11]

When the reaction equation is:

HF ↔ H+ + F-

and when the Ka expression
= concentration of products/concentration of reactions

so, Ka = [H+][F-]/[HF]

when we assume:

[H+] = [F-] = X

and [HF] = 0.35 - X

So, by substitution:

6.8 x 10^-4 = X^2 / (0.35 - X) by solving for X

∴ X = 0.015 M

∴[H+] = X = 0.015

when PH = -㏒[H+]

∴PH = -㏒0.015

= 1.8

6 0

2 years ago