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3 years ago

When liquid gold becomes a solid, its particles get closer together. Which best describes what happened to the gold?

Chemistry

2 answers:

yanalaym [24]3 years ago

7 0

Answer:

The rearrangement of particles in a physical change.

Explanation:

When things are liquid the particles tend to be spread out because they aren't tightly compacted as they would be with a solid. So when liquid gold is changing from a liquid to a solid the properties are changing and the particles in the gold are getting closer together.

Zigmanuir [339]3 years ago

4 0

Answer:

D is correct answer

Explanation:

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Calculate the pH during the titration of 30.00 mL of 0.1000 M HCOOH(aq) with 0.1000 M NaOH(aq) after 29.3 mL of the base have be

pashok25 [27]

Answer:

3.336.

Explanation:

Herein, the no. of millimoles of the acid (HCOOH) is more than that of the base (NaOH).

So, concentration of excess acid = [(NV)acid - (NV)base]/V total = [(30.0 mL)(0.1 M) - (29.3 mL)(0.1 M)]/(59.3 mL) = 1.18 x 10⁻³ M.

For weak acids; [H⁺] = √Ka.C = √(1.8 x 10⁻⁴)(1.18 x 10⁻³ M) = 4.61 x 10⁻⁴ M.

∵ pH = - log[H⁺].

∴ pH = - log(4.61 x 10⁻⁴) = 3.336.

7 0

3 years ago

What did Ernest Rutherford discover? How was his model different from Thomson’s?

s2008m [1.1K]

Answer:

thomson developed the chocolate chip method which was the identification of the electrons in the core of an atom. Rutherford discovered that the core was only positive and that the electrons were floating outside of the core.

Explanation:

4 0

3 years ago

The acid-dissociation constant of hydrocyanic acid (hcn) at 25.0 °c is 4.9 ⋅ 10−10. what is the ph of an aqueous solution of 0.0

vodomira [7]

According to the reaction equation:

and by using ICE table:

CN- + H2O ↔ HCN + OH-

initial 0.08 0 0

change -X +X +X

Equ (0.08-X) X X

so from the equilibrium equation, we can get Ka expression

when Ka = [HCN] [OH-]/[CN-]

when Ka = Kw/Kb

= (1 x 10^-14) / (4.9 x 10^-10)

= 2 x 10^-5

So, by substitution:

2 x 10^-5 = X^2 / (0.08 - X)

X= 0.0013

∴ [OH] = X = 0.0013

∴ POH = -㏒[OH]

= -㏒0.0013

= 2.886

∴ PH = 14 - POH

= 14 - 2.886 = 11.11

5 0

3 years ago

Calculate %ic of the interatomic bonds for the intermetallic compound al6mn. on the basis of this result, what type of interatom

pashok25 [27]

Answer : % ionic character is 0.20%

Bonding between the two metals will be purely metallic.

Explanation: For the calculation of % ionic character, we use the formula

$\% \text{ ionic character}= [1-e^{\frac{-(X_A-X_B)^2)}{4}}]\times(100\%)$

where $X_A$ & $X_B$ are the Pauling's electronegativities.

The table attached has the values of electronegativities, by taking the values of Al and Mn from there,

$X_{Al}=1.61$

$X_{Mn}=1.55$

Putting the values in the electronegativity formula, we get

$\% \text{ ionic character}= [1-e^{\frac{-(1.61-1.55)^2)}{4}}]\times(100\%)$

= 0.20%

Now, there are 3 types of inter atomic bonding

1) Ionic Bonding: It refers to the chemical bond in which there is complete transfer of valence electrons between atoms

2) Covalent Bonding: It refers to the chemical bond involving the sharing of electron pairs between 2 atoms.

3) Metallic Bonding: It refers to the chemical bond in which there is an electrostatic force between the positively charged metal ions and delocalised electrons.

In $Al_6Mn$ compound, the % ionic character is minimal that is $\sim$ 0.20% and there are two metal ions present, therefore this compound will have metallic bonding.

8 0

3 years ago

An a transition metal with 91 protons and electrons?

ser-zykov [4K]

Protactinium

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6 0

3 years ago

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