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g100num [7]
4 years ago
13

Examples for reversible change

Chemistry
2 answers:
castortr0y [4]4 years ago
6 0

Answer:

folding of paper

melting of ice

depletion of ozone

oxidation of blood

Explanation:

Rina8888 [55]4 years ago
4 0
Melting of ice
Sublimation
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50 POINTS WILL MAKE BRAINLIEST!!!!
Bad White [126]

Answer: An example:

Measure the distance between the center of the spreading center (red) and the border between dark yellow and light green (65 Ma point) I measured 2 cm Since the scale on the map is 1 cm = 475 km, calculate the real distance that the plate has moved over the past 65 Ma 2 cm * 475 km/cm = 950 km = 95,000,000 cm = 9.5 * 107 cm Determine the length of time that the plate has been moving65 million years = 65,000,000 years = 6.5 * 107 yearsUse the above equation to calculate spreading rateR = d/t or R = 9.5 * 107 cm / 6.5 * 107 years = 1.46 cm/yr

Explanation:

5 0
2 years ago
What are the boiling points and freezing points (in oC) of a solution of 50.3 g of I2 in 350 g of chloroform? The kb = 3.63 oC/m
patriot [66]

Answer:

Boiling point: 63.3°C

Freezing point: -66.2°C.

Explanation:

The boiling point of a solution increases regard to boiling point of the pure solvent. In the same way, freezing point decreases regard to pure solvent. The equations are:

<em>Boiling point increasing:</em>

ΔT = kb*m*i

<em>Freezing point depression:</em>

ΔT = kf*m*i

ΔT are the °C that change boiling or freezing point.

m is molality of the solution (moles / kg)

And i is Van't Hoff factor (1 for I₂ in chloroform)

Molality of 50.3g of I₂ in 350g of chloroform is:

50.3g * (1mol / 253.8g) = 0.198 moles in 350g = 0.350kg:

0.198 moles / 0.350kg = 0.566m

Replacing:

<em>Boiling point:</em>

ΔT = kb*m*i

ΔT = 3.63°C/m*0.566m*1

ΔT = 2.1°C

As boiling point of pure substance is 61.2°C, boiling point of the solution is:

61.2°C + 2.1°C = 63.3°C

<em>Freezing point:</em>

ΔT = kf*m*i

ΔT = 4.70°C/m*0.566m*1

ΔT = 2.7°C

As freezing point is -63.5°C, the freezing point of the solution is:

-63.5°C - 2.7°C = -66.2°C

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3 years ago
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What is the pH of a 0.222 M acetic acid solution?
Oksi-84 [34.3K]

pH=2.7

<h3>Further explanation</h3>

Acetic acid = weak acid

\tt [H^+]=\sqrt{Ka.M}

Ka = acid ionization constant

M = molarity

Ka for Acetic acid(CH₃COOH) : 1.8 x 10⁻⁵

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