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musickatia [10]
3 years ago
9

What promotes greater hardness in minerals?

Chemistry
2 answers:
Mashutka [201]3 years ago
7 0

Smaller atoms and stronger bonds promotes greater hardness in minerals.

just olya [345]3 years ago
5 0

Answer:

smaller atoms and stronger bonds

Explanation:

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A mothball, composed of naphthalene (c10h8), has a mass of 1.64 g . part a how many naphthalene molecules does it contain?
OLga [1]
The molar mass of Naphthalene is 128g/mol
Therefore; a mass of 1.64 g of Naphthalene contains'
   = 1.64g/128 g
    = 0.0128 moles
But, from the Avogadro's law 1 mole of a substance contains 6.022 × 10^23 particles
Therefore 1 mole of Naphthalene contains 6.022×10^23 molecules
Hence; 0.0128 moles × 6.022 ×10^23 molecules
          = 7.716 × 10^21 molecules
3 0
3 years ago
Are substances that taste bitter and feel slippery to the skin
frez [133]

Answer:

Explanation:

  • Bases are the substances which have a bitter taste and feel slippery.
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4 0
3 years ago
Copper crystallizes in a face-centered cubic lattice (the Cu atoms are at the lattice points and at the face centers). If the de
Delvig [45]

Answer:

The unit cell edge lenght in pm is equal to 361 pm

Explanation:

Data provided:

ρ=Copper density=8.96 g/cm3

Atomic mass of copper=63.54 g/mol

Atoms/cell=4 atoms (in theory)

Avogadro's number=6.02x10^{23} atoms/mol

Since copper has a cubic structure, its cell volume is equal to a^{3}, which can be obtained through the relationship:

cell volume=\frac{(atoms/cell)(atomic mass)}{(density)(Avogadros number)}

Substituting the values:

cell volume=\frac{(4 atoms)(63.54 g/mol)}{(8.96 g/cm3)(6.02x10^{23}) }=4.71x10^{-23}cm^{3}

clearing, we have:

a=\sqrt[3]{4.71x10^{-23}cm^{3}  }=3.61x10^{-8}cm

We convert from centimeter to picometer, 1cm=1x10^{10}pm

a=3.61x10^{-8}cmx\frac{1x10^{10}pm }{1cm} =361 pm

8 0
3 years ago
What is a star, and how do they work
Hitman42 [59]

Answer:

a star is like the sun a burning ball of gas that gives light to the outer space.

Explanation:

8 0
3 years ago
Read 2 more answers
Using the equations
Anna [14]

Considering the Hess's Law, the enthalpy change for the reaction is 221.8 kJ/mol.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

In this case you want to calculate the enthalpy change of:

C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g)

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: H₂ (g) + F₂ (g) → 2 HF (g)     ∆H° = -79.2 kJ/mol

Equation 2: C (s) + 2 F₂ (g) → CF₄ (g)     ∆H° = 141.3 kJ/mol

Equation 3: 2 C(s) + 2 H₂ (g) → C₂H₄ (g)     ∆H° = -97.6 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

<h3 /><h3>FIRST STEP</h3>

First, to obtain the enthalpy of the desired chemical reaction you need one mole of C₂H₄ (g) on reactant side and it is present in first equation. Since this equation has one mole of C₂H₄ (g) on the product side, it is necessary to locate it on the reactant side (invert it).

When an equation is inverted, the sign of ΔH° also changes.

<h3>SECOND STEP</h3>

Now, you need 2 moles of CF₄ (g) on the product side. The second equation has 1 mole of CF₄ (g) on the product side, so it is necessary to multiply it by 2 to obtain 2 moles of CF₄ (g).

Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 2, the variation of enthalpy also.

<h3>THIRD STEP</h3>

Finally, you need 4 moles of  HF (g) on the product side. The first equation has 2 moles of  HF (g) on the product side, so it is necessary to multiply it by 2 to obtain 4 moles of the compound.

Since the equation is multiply by 2, the variation of enthalpy also is multiplied by 2.

<h3>SUMMARY</h3>

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 H₂ (g) + 2 F₂ (g) → 4 HF (g)     ∆H° = -158.4 kJ/mol

Equation 2: 2 C (s) + 4 F₂ (g) → 2 CF₄ (g)     ∆H° = 282.6 kJ/mol

Equation 3: C₂H₄ (g) → 2 C(s) + 2 H₂ (g)     ∆H° = 97.6 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g)     ΔH°= 221.8 kJ/mol

Finally, the enthalpy change for the reaction is 221.8 kJ/mol.

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7 0
2 years ago
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