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ahrayia [7]
3 years ago
14

PLZ HELP!!!! WILL MARK BRAINLYEST!!!!

Chemistry
2 answers:
Simora [160]3 years ago
5 0

the answer is :

1)Earth’s position

3)tectonic activity

4)volcanic activity

5)solar energy

AnnyKZ [126]3 years ago
4 0

Answer:

it's the first,third,fourth,and the fifth.

Explanation:

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Question 1(Multiple Choice Worth 50 points)
irinina [24]

Answer:

because it's shellfish

Explanation:

A slipper

7 0
3 years ago
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Which of the following is a reason to make an armature the parent of a creature
REY [17]

you have to show us the rest of it because we have no idea what your looking at. I'm sorry

3 0
4 years ago
How many moles of NO2 are present in 138.01 g
Sophie [7]

Answer:

3.0002 moles

Explanation:

Data Given:

mass of NO₂ = 138.01 g

No. of moles NO₂ = ?

Solution

To find mass of sample mole formula will be used

mole formula

          no. of mole = mass in grams / molar mass

Molar mass of NO₂ = 14 + 2(16)

Molar mass of NO₂ = 14 + 32 = 46 g/mol

Put values in above equation

               no. of mole = 138.01 g / 46 g/mol

               no. of mole = 3.0002 moles

no. of mole of NO₂ = 3.0002 moles

8 0
3 years ago
Consider the decomposition of red mercury(II) oxide under standard state conditions. )H0 T SFE ڮ( H M 0 H (a) Is the decompositi
sertanlavr [38]

The question is incomplete, complete question is :

Consider the decomposition of red mercury(II) oxide under standard state conditions.

2HgO(s)\rightarrow 2Hg(l)+O_2(g)

Given :

\Delta S_{HgO}^o=70.29 J/mol K

\Delta S_{Hg}^o=75.9 J/mol K

\Delta S_{O_2}^o=205.2 J/mol K

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol

(a) Is the decomposition spontaneous under standard state conditions?

(b) Above what temperature does the reaction become spontaneous?

Answer:

a) The decomposition is spontaneous under standard state conditions.

b)The reaction will spontaneous above -419.69 Kelvins.

Explanation:

2HgO(s)\rightarrow 2Hg(l)+O_2(g)

Given :

\Delta S_{HgO}^o=70.29 J/mol K

\Delta S_{Hg}^o=75.9 J/mol K

\Delta S_{O_2}^o=205.2 J/mol K

Entropy change of the reaction ; ΔS

\Delta S=[2\times \Delta S_{Hg}^o+1\times \Delta S_{O_2}^o]-[2\times \Delta S_{HgO}^o]

=[2\times 75.9 J/mol K+1\times 205.2 J/molK]-[2\times 70.29 J/molK]=216.42 J/mol K

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol  = -90830 J/mol K

1 kJ = 1000 J

At standard condition the value of temperature = T = 298 K

ΔG = ΔH - TΔS

ΔG = -90830 J/mol K - 298 K × 216.42 J/mol K = -155,323.16 J/mol

ΔG < 0 ( spontaneous)

The decomposition is spontaneous under standard state conditions.

b) Above what temperature does the reaction become spontaneous

Let the ΔG = 0

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol  = -90830 J/mol K

ΔG = ΔH - TΔS

0 = -90830 J/mol K - T × 216.42 J/mol K

T = -419.69 K

The reaction will spontaneous above -419.69 Kelvins.

6 0
3 years ago
Copper crystallizes in a face-centered cubic unit cell. the density of copper is 8.94 g/cm3. calculate the length of the edge of
Ilia_Sergeevich [38]
<span>361.4 pm is the length of the edge of the unit cell. First, let's calculate the average volume each atom is taking. Start with calculating how many moles of copper we have in a cubic centimeter by looking up the atomic weight. Atomic weight copper = 63.546 Now divide the mass by the atomic weight, getting 8.94 g / 63.546 g/mol = 0.140685488 mol And multiply by Avogadro's number to get the number of atoms: 0.140685488 * 6.022140857x10^23 = 8.472278233x10^22 Now examine the face-centered cubic unit cell to see how many atoms worth of space it consumes. There is 1 atom at each of the 8 corners and each of those atoms is shared between 8 unit cells for for a space consumption of 8/8 = 1 atom. And there are 6 faces, each with an atom in the center, each of which is shared between 2 unit cells for a space consumption of 6/2 = 3 atoms. So each unit cell consumes as much space as 4 atoms. Let's divide the number of atoms in that cubic centimeter by 4 to determine the number of unit cells in that volume. 8.472278233x10^22 / 4 = 2.118069558x10^22 Now calculate the volume each unit cell occupies. 1 cm^3 / 2.118069558x10^22 = 4.721280262x10^-23 cm^3 Let's get the cube root to get the length of an edge. (4.721280262x10^-23 cm^3)^(1/3) = 3.61426x10^-08 cm Now let's convert from cm to pm. 3.61426x10^-08 cm / 100 cm/m * 1x10^12 pm/m = 361.4 pm Doing an independent search for the Crystallographic Features of Copper, I see that the Lattice Parameter for copper at at 293 K is 3.6147 x 10^-10 m which is in very close agreement with the calculated amount above. And since metals expand and contract with heat and cold, I assume the slight difference in values is due to the density figure given being determined at a temperature lower than 293 K.</span>
4 0
3 years ago
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