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2 years ago

(BRAINLIEST FOR CORRECT ANSWER) How does the particle structure of a solid affect its shape, volume, and compressibility?

Chemistry

1 answer:

Crazy boy [7]2 years ago

6 0

Answer:

A solid's particles fit closely together. The forces between the particles are so strong that the particles can not move freely; they can only vibrate. This causes a solid to be a stable, non-compressible shape with a definite volume.

Explanation:

I honestly don't know if that's right...

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In a titration experiment, hcl and lioh solutions were used. the initial volume of hcl was 1.25ml and lioh was 2.65 ml. the fina

aliya0001 [1]

Answer is: <span>the molarity of HCl is </span>0.097 M.
Chemical reaction: LiOH + HCl → LiCl + H₂O.
V(HCl) = 13.60 mL - 1.25 mL = 12.35 mL.
V(LiOH) = 11.20 mL - 2.65 mL = 8.55 mL.
c(LiOH) = 0.140 M.
From chemical reaction: n(LiOH) : n(HCl) = 1 : 1.
c(HCl) · V(HCl) = c(LiOH) · V(LiOH).
c(HCl) = 8.55 mL · 0.140 M / 12.35 mL.
c(LiOH) = 0.097 M.

8 0

3 years ago

What kind of orbital is occupied by the lone pair electrons in caffeine molecule?

Nataly [62]

The middle nitrogen has two sigma bonds and one pi bond. You know that one p orbital is used in the double bond and two sp2 orbitals are involved in the sigma bond. This leaves one sp2 orbital for the lone pair to occupy.

8 0

3 years ago

A 10.21 mol sample of argon gas is maintained in a 0.7564 L container at 296.9 K. What is the pressure in atm calculated using t

soldi70 [24.7K]

Answer:

The pressure in atm calculated using the van der Waals' equation, is 337.2atm

Explanation:

This is the Van der Waals equation for real gases:

(P + a/v² ) ( v-b) = R .T

where P is pressure

v is Volume/mol

R is the gas constant and T, T° in K

a y b are constant for each gas, so those values are data, from the statement.

[P + 1.345 L²atm/mol² / (0.7564L/10.21mol)² ] (0.7564L/10.21mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K . 296.9K

[P + 1.345 L²atm/mol² / 5.48X10⁻³ L²/mol²] (0.074 L/mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K . 296.9K

(P + 245.05 atm) (0.04181L/mol) = 0.082 L.atm/mol.K . 296.9K

(P + 245.05 atm) (0.04181L/mol) = 24.34 L.atm/mol

0.04181L/mol .P + 10.24 L.atm/mol = 24.34 L.atm/mol

0.04181L/mol .P = 24.34 L.atm/mol - 10.24 L.atm/mol

0.04181L/mol. P = 14.1 L.atm/mol

P = 14.1 L.atm/mol / 0.04181 mol/L

P = 337.2 atm

4 0

2 years ago

A balance is used to measure: mass weight volume density

Crank

A balance is used to determine mass

8 0

3 years ago

Read 2 more answers

Sodium phosphate is added to a solution that contains 0.0030 M aluminum nitrate and 0.016 M calcium chloride. The concentration

gavmur [86]

Explanation:

It is given that aluminium nitrate and calcium chloride are mixed together with sodium phosphate.

And, $K_{sp} of AlPO_{4} = 9.84 \times 10^{-21}$

$K_{sp} of Ca_{3}(PO_{4})_{2} = 2.0 \times 10^{-29}$

Let us assume that the solubility be "s". And, the reaction equation is as follows.

$AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}$

$9.84 \times 10^{-21} = s \times s$

s = $9.92 \times 10^{-11}$

Also, $Ca_{3}(PO_{4})_{2} \rightleftharpoons 3Ca^{2+} + 2PO^{3-}_{4}$

$2 \times 10^{-29} = (3s)^{3} \times (2s)^{2}$

s = $7.14 \times 10^{-7}$

This means that first, aluminium phosphate will precipitate.

Now, we will calculate the concentration of phosphate when calcium phosphate starts to precipitate out using the $K_{sp}$ expression as follows.

$K_{sp} = [Ca^{2+}]^{3}[PO^{3-}_{4}]^{2}$

$2.0 \times 10^{-29} = (0.016)^{3}[PO^{3-}_{4}]^{2}$

$2.0 \times 10^{-29} = 4.096 \times 10^{-6} \times [PO^{3-}_{4}]^{2}$

$[PO^{3-}_{4}]^{2}$ = $4.88 \times 10^{-24}$

= $2.21 \times 10^{-12}$ M

Similarly, calculate the concentration of aluminium at this concentration of phosphate as follows.

$AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}$

$K_{sp} = [Al^{3+}][PO^{3-}_{4}]$

$9.84 \times 10^{-21} = [Al^{3+}] \times 2.21 \times 10^{-12}$

$[Al^{3+}] = 4.45 \times 10^{-9}$ M

Thus, we can conclude that concentration of aluminium will be $4.45 \times 10^{-9}$ M when calcium begins to precipitate.

7 0

2 years ago