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2 years ago

For the reaction Cl2 + 2KBr → 2KCl + Br2, how many moles of potassium chloride are produced from 4.3 moles of chlorine?

1 answer:

8 0

<span>Cl</span>₂<span> + 2 KBr = 2 KCl + Br</span>₂
↓ ↓
1 mole Cl₂ ------ 2 mole KCl
4.3 moles Cl₂---- ?

Moles KCl = 4.3 * 2 / 1

= 8.6 moles KCl

hope this helps!

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2 years ago

A 1.00 cm2 blackbody surface emits 201 watts of radiant power. What is the temperature of the surface? Give to the nearest whole

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Explanation:

Formula for black body radiation is as follows.

$\frac{P}{A} = \sigma \times T^{4} J/m^{2}s$

where, P = power emitted

A = surface area of black body

$\sigma$ = Stephen's constant = $5.6703 \times 10^{-8} watt/m^{2}.K^{-4}$

As area is given as 1.0 $cm^{2}$ . Converting it into meters as follows.

$1.00 cm^{2} \times \frac{(10^{-2})^{2} m^{2}}{1 cm^{2}}$ (as 1 m = 100 cm)

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It is given that P = 201 watts. Hence,

$\frac{201 watts}{1 \times 10^{-4} m^{2}}$ = $5.6703 \times 10^{-8} watt/m^{2}.K^{-4} \times T^{4}$

$T^{4}$ = $35.45 \times 10^{12}$

T = $(35.45 \times 10^{12})^{1/4}$

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Thus, we can conclude that the temperature of the surface is 8862.5 K.

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2 years ago

Which structure of the organism work together to carry out all of its life functins?

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2 years ago

When 70.4 g of benzamide (C7H7NO) are dissolved in 850. g of a certain mystery liquid X, the freezing point of the solution is 2

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Answer:

1.62

Explanation:

From the given information:

number of moles of benzamide $=\dfrac{70.4 \ g}{121.14 \ g/mol}$

= 0.58 mole

The molality = $\dfrac{mass \ of \ solute (i.e. \ benzamide )}{mass \ of \ solvent }$

$= \dfrac{0.58 }{0.85 }$

= 0.6837

Using the formula:

$\mathbf {dT = l \times k_f \times m}$

where;

dT = freezing point = 27

l = Van't Hoff factor = 1

kf = freezing constant of the solvent

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2.7 °C = 1 × kf × 0.6837 m

kf = 2.7 °C/ 0.6837m

kf = 3.949 °C/m

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= 1.316 mol

The molality = $\dfrac{1.316 \ mol}{0.85 \ kg}$

= 1.5484

Thus;

the above kf value is used in determining the Van't Hoff factor for NH4Cl

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9.9 = l × 3.949 × 1.5484 m

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2 years ago

What is the product if an atom of Po-209<br> undergoes alpha decay?

g100num [7]

<h3>Answer:</h3>

Lead-205 (Pb-205)

<h3>Explanation:</h3>

<u>We are given;</u>

An atom of Po-209

We are supposed to identify its product after an alpha decay;

Polonium-209 has a mass number of 209 and an atomic number of 84.
When an element undergoes an alpha decay, the mass number decreases by 4 while the atomic number decreases by 2.
Therefore, when Po-209 undergoes alpha decay it results to the formation of a product with a mass number of 205 and atomic number of 82.
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The equation for the decay is;

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Note; An alpha particle is represented by a helium nucleus, ⁴₂He.

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