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3 years ago

At room temperature, which of these are oxide minerals? Select all that apply.

Chemistry

2 answers:

DerKrebs [107]3 years ago

7 0

C. and d. are oxyminerals.

Klio2033 [76]3 years ago

3 0

Answer: C. $Al_{2}O_{3}$ and D. $Fe_{3}O_{4}$

Explanation: Oxide minerals are the minerals in which the oxide anion is bonded to the central metal atom. These are useful for industrial purposes.

Out of the given options, only two of them can be categorized as the oxide minerals.

H2O and CO2 will not be considered as oxide minerals.

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What animals use the earths magnetic field for navigation

Ivenika [448]

Birds use the magnetic fields and some fish i believe as well and maybe butterflies im not entirely sure

6 0

3 years ago

Antarctica, almost completely covered in ice, has an area

Marysya12 [62]

Answer: The mass of ice is $2.39\times 10^{22}g$

Explanation:

We are given:

Area of Antarctica = $5,500,000mi^2=5,500,000\times 2.59\times 10^{10}=142.45\times 10^{15}cm^2$ (Conversion factor: $1mi^2=2.59\times 10^{10}cm^2$ )

Height of Antarctica with ice = 7500 ft.

Height of Antarctica without ice = 1500 ft.

Height of ice = 7500 - 1500 = 6000 ft = $182.88\times 10^3cm$ (Conversion factor: 1 ft = 30.48 cm)

To calculate mass of ice, we use the equation:

$\text{Density of a substance}=\frac{\text{Mass of a substance}}{\text{Volume of a substance}}$

We are given:

Density of ice = $0.917g/cm^3$

Volume of ice = Area × Height of ice = $142.45\times 10^{15}cm^2\times 182.88\times 10^3cm=26051.26\times 10^{18}cm^3$

Putting values in above equation, we get:

$0.917g/cm^3=\frac{\text{Mass of ice}}{26051.26\times 10^{18}cm^3}\\\\\text{Mass of ice}=(0.917g/cm^3\times 26051.26\times 10^{18}cm^3=2.39\times 10^{22}g$

Hence, the mass of ice is $2.39\times 10^{22}g$

5 0

3 years ago

(WILL MARK BRAINLIEST) PLZ HELP ASAP

ArbitrLikvidat [17]

Answer:

a) 2NaOH(aq) + CuSO4(aq) -------------> Cu(OH)2(s) + Na2SO4(aq)

b) Ca(OH)2(aq) + CO2(g) --------------> CaCO3 + H2O (this is already balanced)

c) Pb(NO3)2 + H2SO4 --------> PbSO4 + 2HNO3.

d) 2KNO3 ------> 2KNO2 + O2

e) H2SO4 + 2(NaOH) -----> Na2SO4 + 2(H2O)

f) Ca(NO3)2(aq) + (NH4)2CO3(aq) ----------------> CaCO3(s) + 2NH4NO3(aq)

3 0

3 years ago

Read 2 more answers

63Ni decays by a first-order process via the emission of a beta particle. The 63Ni isotope has a half-life of 100. years. How lo

rewona [7]

Answer:

151.4863 years

Explanation:

Half life, t1/2 = 100 years

Initial concentration,[A]o = 100%

Final concentration, [A] = 35% (after 65% have been decayed)

Time = ?

Half life for a first Order reaction is given as;

t1/2 = ln (2) / k

k = ln(2) / 100

k = 0.00693y-1

The integral rate law for first order reactions is given as;

ln[A] = ln[A]o − kt

kt = ln[A]o - ln[A]

t = ( ln[A]o - ln[A]) / k

t = [ln(100) - ln(35)] /0.00693

t = 1.0498 / 0.00693

t = 151.4863 years

8 0

3 years ago

A plot of binding energy per nucleon (Eb/ A) versus the mass number (A) shows that nuclei with a small mass number have a small

juin [17]

Answer:

a) 1.12 MeV / nucleon

b) 5.62 MeV / nucleon

c) 8.80 MeV / nucleon

d) 8.56 MeV / nucleon

we can conclude that the binding energy has a maximum value for nuclei with a mass around 60

Explanation:

Binding energy = ( Δm * 931.5 ) MeV

Binding energy per nucleon = Binding energy in / Number of nucleon

a) ²H = 1 neutron , 1 proton = 2 nucleons

Given that the theoretical mass = 2.0141 u

Actual mass = 1.0078 u + 1.0087 u = 2.0165 u

Δm = 2.0165 u - 2.0141 u = 2.4 * 10^-3 u

∴ Binding energy per nucleon = ( 2.4 * 10^-3 * 931.5 ) MeV / 2 nucleons

= 1.12 MeV / nucleon

b) ⁷Li = 3 protons , 4 neutrons = 7 nucleons

theoretical mass = 7.0160 u

Actual mass = ( 3 * 1.0078 ) + ( 4 * 1.0087 ) = 7.0582 u

Δm = ( 7.0582 u - 7.0160 u ) = 0.0422 u

∴ Binding energy per nucleon = ( 0.0422 * 931.5 ) / 7

= 5.62 MeV / nucleon

C) ⁶²Ni = 28 protons , 34 neutrons = 62 nucleons

Theoretical mass = 61.9283 u

Actual mass = ( 28 * 1.0078 ) u + ( 34 * 1.0087 ) u

= 62.5142 u

Δm = 0.5859 u

∴ Binding energy per nucleon = ( 0.5859 * 931.5 ) / 62

= 8.80 MeV / nucleon

D) ¹¹⁰Cd = 48 protons , 62 neutrons = 110 nucleons

Theoretical mass = 109.9030 u

Actual mass = ( 48 * 1.0078 ) + ( 62 * 1.0087 )

= 110.9138 u

Δm = ( 110.9138 - 109.9030 ) = 1.0108 u

∴ Binding energy per nucleon = ( 1.0108 * 931.5 ) / 110

= 8.56 MeV / nucleon

hence we can conclude that the binding energy has a maximum value for nuclei with a mass around 60

3 0

3 years ago