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3 years ago

Select all that apply to the property of hardness.

Chemistry

2 answers:

umka21 [38]3 years ago

5 0

Answer:

hardness is a physical property

hardness is indicative of the strength of chemical bonds between elements.

hardness is a determined through a complicated processs that requires expensive instruments.

Friedrish Mohs developed the standard hardness scale.

Diamomd can scratch quartz

boyakko [2]3 years ago

4 0

Answer:

Hardness is a physical property

Friedrich Mohs developed the standard hardness scale.

Diamond can scratch quartz.

Explanation:

hope it helped! ^_^

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A sample of gas at 313 K and 4.95 atm occupies 85.62 L of space. Which

Daniel [21]

Answer: 85.62 L represents volume (V).

Explanation:

Volume is defined as the amount or quantity of a given substance or an object.

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 4.95 atm

V = Volume of gas = 85.62 L

n = number of moles

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $313K$

Thus 85.62 L represents volume (V).

8 0

4 years ago

2.3 Zinc has five naturally occurring isotopes: 48.63% of 64 Zn with an atomic weight of 63.929 amu; 27.90% of 66Zn with an atom

lakkis [162]

Answer: The average atomic mass of element Zinc is 65.40 amu.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

For $_{30}^{64}\textrm{Zn}$ isotope:

Mass of $_{30}^{64}\textrm{Zn}$ isotope = 63.929 amu

Percentage abundance of $_{30}^{64}\textrm{Zn}$ isotope = 48.63 %

Fractional abundance of $_{30}^{64}\textrm{Zn}$ isotope = 0.4863

For $_{30}^{66}\textrm{Zn}$ isotope:

Mass of $_{30}^{66}\textrm{Zn}$ isotope = 65.926 amu

Percentage abundance of $_{30}^{66}\textrm{Zn}$ isotope = 27.90 %

Fractional abundance of $_{30}^{66}\textrm{Zn}$ isotope = 0.2790

For $_{30}^{67}\textrm{Zn}$ isotope:

Mass of $_{30}^{67}\textrm{Zn}$ isotope = 66.927 amu

Percentage abundance of $_{30}^{67}\textrm{Zn}$ isotope = 4.10 %

Fractional abundance of $_{30}^{67}\textrm{Zn}$ isotope = 0.0410

For $_{30}^{68}\textrm{Zn}$ isotope:

Mass of $_{30}^{68}\textrm{Zn}$ isotope = 67.925 amu

Percentage abundance of $_{30}^{68}\textrm{Zn}$ isotope = 18.75 %

Fractional abundance of $_{30}^{68}\textrm{Zn}$ isotope = 0.1875

For $_{30}^{70}\textrm{Zn}$ isotope:

Mass of $_{30}^{70}\textrm{Zn}$ isotope = 69.925 amu

Percentage abundance of $_{30}^{70}\textrm{Zn}$ isotope = 0.62 %

Fractional abundance of $_{30}^{70}\textrm{Zn}$ isotope = 0.0062

Putting values in equation 1, we get:

$\text{Average atomic mass of Zinc}=[(63.929\times 0.4863)+(65.926\times 0.2790)+(66.927\times 0.0410)+(67.925\times 0.1875)+(69.925\times 0.0062)]$

$\text{Average atomic mass of Zinc}=65.40amu$

Hence, the average atomic mass of element Zinc is 65.40 amu.

7 0

4 years ago

In certain conditions, the equilibrium concentrations for components in the reaction 2NOCl(g) 2NO(g) + Cl2(g) are [NO] = 0.02 M,

Sonja [21]

Keq= (products)/ (reactants)

Keq= ( [NO]^2 x [Cl2]) / ( [NOCl]^2)

Keq= ( (0.02)^2 x (0.01) ) / (0.5)^2= 1.6 x 10-5

8 0

3 years ago

12.5 g of copper are reacted with an excess of chlorine gas, and 25.4 g of copper(II) chloride are

Alika [10]

Answer:

Percent yield = 94.5%

Theoretical yield = 26.89 g

Explanation:

Given data:

Mass of copper = 12.5 g

Mass of copper chloride produced = 25.4 g

Theoretical yield = ?

Percent yield = ?

Solution:

Cu + Cl₂ → CuCl₂

Number of moles of Copper:

Number of moles = mass/ molar mass

Number of moles = 12.5 g/ 63.55 g/mol

Number of moles = 0.2 mol

Now we will compare the moles of copper with copper chloride.

Cu : CuCl₂

1 : 1

0.2 : 0.2

Theoretical yield:

Mass of copper chloride:

Mass = Number of moles × molar mass

Mass = 0.2 mol × 134.45 g/mol

Mass = 26.89 g

Percent yield:

Percent yield = Actual yield / theoretical yield × 100

Percent yield = 25.4 g/26.89 g × 100

Percent yield = 94.5%

8 0

4 years ago

Calculate the number of formula units in 2.50 mol NaNO

marta [7]

The number of formula units in 2.50 mol of the compound is 15.1 * 10^23.

The question is unclear whether NaNO2 or NaNO3 is implied. However,in either case, the solution applies equally.

6.02 * 10^23 formula units of the compound are contained in 1 mole

x formula units are contained in 2.5 moles of the compound

x = 6.02 * 10^23 formula units * 2.5 moles/ 1 mole

x = 15.1 * 10^23 formula units of the compound.

Learn more; brainly.com/question/9743981

3 0

3 years ago