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3 years ago

Which is one factor of rock types that affects the rate of weathering?

Chemistry

2 answers:

avanturin [10]3 years ago

7 0

Answer:

Its A porosity

Explanation:

Luba_88 [7]3 years ago

4 0

Answer:

Explanation:

got it right on edge

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When the vapor pressure of a liquid is equal to the atmospheric pressure, the liquid

Dmitrij [34]

Answer:

The liquid will boil.

Explanation:

Boiling will occur when the vapor pressure is equal to the atmospheric pressure. This is called the boiling point. Without any external pressure the liquid molecules will be able to spread out and change from a liquid to a gas.

7 0

3 years ago

En que flota el osmio

Valentin [98]

Translate to english

6 0

3 years ago

A compound is found to contain 42.88 % carbon and 57.12 % oxygen by weight. To answer the questions, enter the elements in the o

Ipatiy [6.2K]

Answer:

The empirical formule is CO

Explanation:

Step 1: Data given

Suppose the mass of a compound is 100 grams

Suppose the compound contains:

42.88 % C = 42.88 grams C

57.12 % O = 57.12 grams O

Molar mass C = 12.01 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles C = mass C / molar mass C

Moles C = 42.88 grams / 12.01 g/mol

Moles C = 3.57 moles

Moles O = 57.12 grams / 16.0 g/mol

Moles O = 3.57 moles

Step 3: Calculate the mol ratio

We divide by the smallest amount of moles

C: 3.57 moles / 3.57 moles = 1

O: 3.57 moles / 3.57 moles = 1

The empirical formule is CO

3 0

3 years ago

The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling poi

Gelneren [198K]

Answer : The normal boiling point of ethanol will be, $348.67K$ or $75.67^oC$

Explanation :

The Clausius- Clapeyron equation is :

$\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = vapor pressure of ethanol at $30^oC$ = 98.5 mmHg

$P_2$ = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

$T_1$ = temperature of ethanol = $30^oC=273+30=303K$

$T_2$ = normal boiling point of ethanol = ?

$\Delta H_{vap}$ = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

$\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})$

$T_2=348.67K=348.67-273=75.67^oC$

Hence, the normal boiling point of ethanol will be, $348.67K$ or $75.67^oC$

3 0

3 years ago

Name this compound [TiBr3(NMe3)3]

madreJ [45]

<span> Titanium tribromide, titanium (III) bromide, or titanous bromide. </span>

5 0

3 years ago

Read 2 more answers