14 point. Please help need this ASAP.

A) Explain what is meant by anisotropic refinement <br>

Rina8888 [55]

Answer:

ok

Explanation:

In anisotropic refinement, this motion is fitted as an ellipsoid, the center of which lies inside the circle of motion, while the atom is loated on its periphery. This effect makes the bond distance appear shorter.

(answer is from go0gle not gonna say its lie)

6 0

2 years ago

To make use of an ionic hydrate for storing solar energy, you place 422.0 kg of sodium sulfate decahydrate on your house roof. A

Law Incorporation [45]

Complete Question

To make use of an ionic hydrate for storing solar energy, you place 422.0 kg of sodium sulfate decahydrate on your house roof. Assuming complete reaction and 100% efficiency of heat transfer, how much heat (in kJ) is released to your house at night? Note that sodium sulfate decahydrate will transfer 354 kJ/mol.

Answer:

The amount of energy released is $x = 4.63650 *10^5 KJ$

Explanation:

Number of moles is mathematically represented as

$n =\frac{mass}{Molar\ Mass}$

substituting $422.0kg = 422 *10^3g$ for mass of sodium sulfate decahydrate( $Na_2 SO_4 \cdot 10H_2 O$ ), $322.2g/mol$ (This value is a constant )for the molar mass of sodium sulfate decahydrate

$n= \frac{422*10^3}{322.2}$

$= 1309.7 \ moles$

From the question we are told that

1 mole of sodium sulfate decahydrate generates $354KJ$ of energy

So 1309.7 mole would generate x

Now stating the relation mathematically

1 mol → 354KJ

1309.7 mol → x

=> $x = 4.63650 *10^5 KJ$

5 0

3 years ago

Citric acid is a naturally occurring compound. what orbitals are used to form each indicated bond? be sure to answer all parts.2

Rina8888 [55]

Answer : The orbitals that are used to form each indicated bond in citric acid is given below as per the attachment.

Answer 1) σ Bond a : C has $SP^{2}$ O has $SP^{2}$ .

Explanation : The orbitals of oxygen and carbon which are involved in $SP^{2}$ hybridization to form sigma bonds. This is observed at 'a' position in the citric acid molecule.

Answer 2) π Bond a: C has π orbitals and O also has π orbitals.

Explanation : The pi-bond at the 'a'position has carbon and oxygen atoms which undergoes pi-bond formation. And has pi orbitals of oxygen and carbon involved in the bonding process.

Answer 3) Bond b: O $SP^{3}$ H has only S orbital involved in bonding.

Explanation : The bonding at 'b' position involves oxygen $SP^{3}$ hydrogen atoms in it. It has $SP^{3}$ hybridized orbitals and S orbital of hydrogen involved in the bonding.

Answer 4) Bond c: C is $SP^{3}$ O is also $SP^{3}$

Explanation : The bonding involved at 'c' position has carbon and oxygen atoms involved in it. Both the atoms involves the orbitals of $SP^{3}$ hybridized bonds.

Answer 5) Bond d: C atom has $SP^{3}$ C atom has $SP^{3}$

Explanation : At the position of 'd' the bonding between two carbon atoms is found to be $SP^{3}$ . Therefore, the orbitals that undergo $SP^{3}$ hybridization are $SP^{3}$ .

Answer 6) Bond e : C1 containing O $SP^{2}$

C2 is $SP^{3}$

Explanation : The carbon atom which contains oxygen along with a double bond has $SP^{2}$ hybridized orbitals involved in the bonding process; whereas the carbon at C2 has $SP^{3}$ hybridized orbitals involved during the bonding. This is for the 'e' position.