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2 years ago

Chalk markings cannot be easily rubbed off from a black board if kept for a long time.why?

Chemistry

1 answer:

snow_tiger [21]2 years ago

3 0

The chalk particles embed themselves into the small pores on the surface.

Although a chalkboard seems smooth to the touch, it is quite rough at the microscopic level, with pores that reach below the surface.

When you drag chalk across the board, friction causes small particles of chalk to rub off onto the surface.

If you leave the markings for a long time, some of the chalk particles will work their way into the pores.

A brush will remove the surface particles, but it will not be able to get at the particles in the pores.

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HELIUM) has chemical properties so different from those of the other elements that it cannot be placed in any group. *

siniylev [52]

Answer:

jesus

Explanation:

jesus is always the answer

6 0

2 years ago

The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions

mamaluj [8]

Answer:

a) 62.1 kJ/mol

b) 2.82 kJ/mol

c) 270.91 kJ/mol

d) -851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

a) 2Ag₂O(s) → 4Ag(s) + O₂(g)

ΔH°f, Ag₂O(s) = -31.05 kJ/mol

ΔH°rxn = 0 - (2*(-31.05)) = 62.1 kJ/mol

b) SnO(s) + CO(g) → Sn(s) + CO₂(g)

ΔH°f,SnO(s) = -285.8 kJ/mol

ΔH°f,CO(g) = -110.53 kJ/mol

ΔH°f,CO₂(g) = -393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l)

ΔH°f,Cr₂O₃(s) = -1128.4 kJ/mol

ΔH°f,H₂O(l) = -285.83 kJ/mol

ΔH°rxn = [3*(-285.83)] - [( -1128.4)] = 270.91 kJ/mol

d) 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

ΔH°f,Fe₂O₃(s) = -824.2 kJ/mol

ΔH°f,Al₂O₃(s) = -1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = -851.5 kJ/mol

3 0

3 years ago

How many atoms are in 10.1 g Ne

LekaFEV [45]

Answer:

3.01 × 10²³ atoms Ne

General Formulas and Concepts:

Atomic Structure

Reading a Periodic Tables
Moles

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

Identify

[Given] 10.1 g Ne

[Solve] atoms Ne

Step 2: Identify Conversions

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

[PT] Molar Mass of Ne: 20.18 g/mol\

Step 3: Convert

[DA] Set up: $\displaystyle 10.1 \ g \ Ne(\frac{1 \ mol \ Ne}{20.18 \ g \ Ne})(\frac{6.022 \cdot 10^{23} \ atoms \ Ne}{1 \ mol \ Ne})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 3.01398 \cdot 10^{23} \ atoms \ Ne$