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Nastasia [14]
3 years ago
12

Which mineral has the same hardness as a fingernail?

Chemistry
2 answers:
Marina86 [1]3 years ago
4 0
Gypsum has the same hardness as a fingernail
ipn [44]3 years ago
4 0
<h3>Answer:</h3>

                  Gypsum

<h3>Explanation:</h3>

                   The hardness of minerals or substances is measured qualitatively using Mohs Scale. In this scale the hardness is measured qualitatively by characterizing the resistance shown by one substance when rubbed against second substance. Therefore, greater points are given to those substances which are capable of making scratches on other substances. The Mohs Scale of given compounds is as follow,

                                     Diamond        <em>  10</em>

                                     Fluorite          <em>  4</em>

                                      Topaz                <em>8</em>

                                      Gypsum            <em>1.5 to 2</em>

While the hardness value of Fingernail is <em>2 to 2.5</em>. Hence, this value is close to Mohs scale hardness value of Gypsum therefore, gypsum is the correct answer.

You might be interested in
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
Given that the ka for hocl is 3.5 × 10–8, calculate the k value for the reaction of hocl with oh–.
Harlamova29_29 [7]
Following reaction is involved in above system
HOCl(aq)  ↔  H+(aq) + OCl-<span>(aq)
</span>OCl-(aq) + H2O(l)  ↔ HOCl(aq) + OH-<span>(aq)
</span>
Now, if the system is obeys 1st order kinetics we have
K  = [OCl-][H+<span>]/[HOCl]   ............. (1)
</span>∴ [HOCl-] / [OCl-] = [H+] (1 / 3.0 * 10-8<span>)   ............. (2)
</span>
and now considering that system is obeying 2nd order kinetics, we have
K  = [OH-][HOCl-] / [OCl-] ................. (3<span>)
</span>Subs 2 in 3 we get
K = [OH-][H+] (1 / 3.0 * 10-8<span>)
</span>we know that, [OH-][H+] = 10<span>-14

</span>∴K = 3.3 * 10<span>-7
</span>
Thus, correct answer is e i.e none of these
6 0
3 years ago
Read 2 more answers
5. How can you tell the difference between CuS and Cu2S
puteri [66]

Answer:

Density, melting point. and magnetic properties  

Explanation:

I can think of three ways.

1. Density

The density of Cu₂S is 5.6 g/cm³; that of CuS is 4.76 g/cm³.

It should be possible to distinguish these even with high school equipment.

2. Melting point

Cu₂S melts at 1130 °C (yellowish-red); CuS decomposes at 500 °C (faint red).

A Bunsen burner can easily reach these temperatures.

3. Magnetic properties

You can use a Gouy balance to measure the magnetic susceptibilities.

In Cu₂S the Cu⁺ ion has a d¹⁰ electron configuration, so all the electrons are paired and the solid is diamagnetic.

In CuS the Cu²⁺ ion has a d⁹ electron configuration, so all there is an unpaired electron and the solid is paramagnetic.

A sample of Cu₂S will be repelled by the magnetic field and show a decrease in weight.

A sample of CuS will be attracted by the magnetic field and show an increase in weight.

In the picture below, you can see the sample partially suspended between the poles of an electromagnet.

 

6 0
3 years ago
What mass of C6H8O7 should be used every 7.0 X 10^2mg NaHCO3
Savatey [412]

Mass C₆H₈O₇ : 0.531484 g

<h3>Further explanation</h3>

Reaction

3NaHCO₃ (aq) + C₆H₈O₇ (aq) → 3 CO₂ (g) + 3 H₂O (l) + Na₃C₆H₅O₇ (aq)

MW NaHCO₃ : 84 g/mol

mass NaHCO₃ : 7.10² mg=0.7 g

mol NaHCO₃ :

\tt mol=\dfrac{0.7}{84}=0.0083

mol C₆H₈O₇ :

\tt \dfrac{1}{3}\times 0.0083=0.00277

MW C₆H₈O₇ : 192 g/mol

mass C₆H₈O₇ :

\tt mass=0.00277\times 192=0.53184

4 0
3 years ago
Pressure is this sample?
faltersainse [42]

Answer:

P = 164 Atm

Explanation:

PV = nRT => P = nRT/V

n = 10.0 moles

R = 0.08206 L·Atm/mol·K

T = 27.0°C = 300 K

V = 1.50 Liters

P = (10.0 mol)(0.08206 L·Atm/mol·K )(300 K)/(1.50 Liters) = 164.12 Atm ≅ 164 Atm (3 sig. figs.)

5 0
3 years ago
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