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tresset_1 [31]
3 years ago
8

List 2 extensive properties and 3 intensive properties of a 5.0g, 1 cm3 cube of silver

Chemistry
1 answer:
jeka57 [31]3 years ago
3 0

Properties of matter can be broadly classified into two categories:

Physical properties which usually involve a change in the state of matter and Chemical properties which involve a change in the chemical composition of matter.

Now, physical properties can be further classified as:

Extensive: these depend on the amount of the substance, eg: mass, volume  

Intensive: these do not depend on the amount of the substance eg: density, color, melting point, boiling point

Here we are given a 5.0 g and 1 cm3 silver cube :

Therefore:

Extensive properties are-

1) Mass of silver = 5.0 g

2) Volume of silver = 1 cm3

Intensive properties are:

1) Density of silver = mass/volume = 5.0 g/ 1 cm3 = 5.0 g/cm3

2) Melting point of silver = 962 C

3) Color = white/gray

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PolarNik [594]

Answer:

Electromagnetic Waves

Explanation:

Electromagnetic waves do not need a medium to travel through, so therefore, they are faster.

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The Dome of the Rock and Al-Aqsa Mosque are located in what city?
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They are located in Jerusalem.
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Describe the buffer capacity of the acetic acid buffer solution in relation to the addition of both concentrated and dilute acid
AysviL [449]

Answer:

The answer is in the explanation

Explanation:

Acetic acid, CH₃COOH, is a weak acid that will produce a buffer when its conjugate base, CH₃COO⁻, acetate ion, is added to the solution.

<em>That is because a buffer is the mixture of a weak acid and its conjugate base or vice versa.</em>

When an acid (HX) is added to the solution, the acetate ion will react producing acetic acid, thus:

CH₃COO⁻ + HX → CH₃COOH + X⁻

For this reason, the pH doesn't change abruptly because H⁺ ions are not produced.

Now, if a  base (BOH) is added to the buffer, CH₃COOH will react producing acetate ion and water, thus:

CH₃COOH + BOH → CH₃COO⁻ + H₂O + B⁺.

In the same way, there are not produced free OH⁻ and the pH doesn't change significantly.

5 0
3 years ago
During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.
Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

  • CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) /  (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) /  (32.00 g/mol) = 0.2812 mol.

  • it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of  CH₄ needs → 2 mol of O₂

???  mol of  CH₄  needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over *  molar mass

                                    = 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0
3 years ago
In acidic aqueous solution, the purple complex ion Co(NH3)5Br2+ undergoes a slow reaction in which the bromide ion is replaced b
kirill115 [55]

Answer:

A) 0.065 M is its molarity after a reaction time of 19.0 hour.

B) In 52 hours [Co(NH_3)5Br]^{2+} will react 69% of its initial concentration.

Explanation:

Co(NH_3)_5(H_2O)_3+[Co(NH_3)5Br]^{2+}(Purple)(aq)+H_2O(l)\rightarrow [Co(NH_3)_5(H_2O)]^{3+}(Pinkish-orange)(aq)+Br^-(aq)

The reaction is first order in [Co(NH_3)5Br]^{2+}:

Initial concentration of [Co(NH_3)5Br]^{2+}= [A_o]=0.100 M

a) Final concentration of [Co(NH_3)5Br]^{2+} after 19.0 hours= [A]

t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)

Rate constant of the reaction = k = 6.3\times 10^{-6} s^{-1}

The integrated law of first order kinetic is given as:

[A]=[A_o]\times e^{-kt}

[A]=0.100 M\times e^{-6.3\times 10^{-6} s^{-1}\times 19.0\times 3600 s}

[A]=0.065 M

0.065 M is its molarity after a reaction time of 19.0 h.

b)

Initial concentration of [Co(NH_3)5Br]^{2+}= [A_o]=x

Final concentration of [Co(NH_3)5Br]^{2+} after t = [A]=(100\%-69\%) x=31\%x=0.31x

Rate constant of the reaction = k = 6.3\times 10^{-6} s^{-1}

The integrated law of first order kinetic is given as:

[A]=[A_o]\times e^{-kt}

0.31x=x\times e^{-6.3\times 10^{-6} s^{-1}\t}

t = 185,902.06 s = \frac{185,902.06 }{3600} hour = 51.64 hours ≈ 52 hours

In 52 hours [Co(NH_3)5Br]^{2+} will react 69% of its initial concentration.

6 0
3 years ago
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