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3 years ago

Which term correctly describes all of the following materials ? Salt water,vinegar , bronze,air,beach sand?

Chemistry

2 answers:

maksim [4K]3 years ago

7 0

Answer:

ummmm hey

Explanation:

du sus su du du su du

Sphinxa [80]3 years ago

4 0

They are all mixtures so yea

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Expain why 4-methyl pentane is not a correct name

ziro4ka [17]

Answer: 2-Methylpentane

Explanation: The numbering is not brought to a minimum.

8 0

3 years ago

Phosphorus crystallizes in several different forms, one of which is a simple cube with an edge length of 238 pm. What is the den

algol13

Answer: The density of phosphorus is $3.81g/cm^3$

Explanation:

To calculate the density of phosphorus, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density

Z = number of atom in unit cell = 1 (CCP)

M = atomic mass of phosphorus = 31 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell = $238pm=238\times 10^{-10}cm$ (Conversion factor: $1cm=10^{10}pm$ )

Putting values in above equation, we get:

$\rho=\frac{1\times 31}{6.022\times 10^{23}\times (238\times 10^{-10})^3}\\\\\rho=3.81g/cm^3$

Hence, the density of phosphorus is $3.81g/cm^3$

6 0

3 years ago

Name two objects which are colourless liquid

mina [271]

Answer:

water and asid sulfuric

4 0

3 years ago

Read 2 more answers

The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av

nlexa [21]

Answer: The percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

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)

Let the fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope be 'x'. So, fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope will be '1 - x'

For $_{77}^{191}\textrm{Ir}$ isotope:

Mass of $_{77}^{191}\textrm{Ir}$ isotope = 190.9606 amu

Fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope = x

For $_{77}^{193}\textrm{Ir}$ isotope:

Mass of $_{77}^{193}\textrm{Ir}$ isotope = 192.9629 amu

Fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

$192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710$

Percentage abundance of $_{77}^{191}\textrm{Ir}$ isotope = $0.3710\times 100=37.10\%$

Percentage abundance of $_{77}^{193}\textrm{Ir}$ isotope = $(1-0.3710)=0.6290\times 100=62.90\%$

Hence, the percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

8 0

3 years ago

HELPHELPHELPHELPHELPHELPHELPHELPHELP

aalyn [17]

the answer is c.50atm

Explanation:

xddgghbchjudsxbnkkhdxegjbch

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3 years ago