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3 years ago

One sentence describing the difference between pure substances and mixtures.

Chemistry

2 answers:

Alja [10]3 years ago

6 0

Answer:

A pure substance is made up of only one ingredient or combination. A mixture is made up of two or maybe more separate components that are not chemically linked.

Explanation:

aev [14]3 years ago

5 0

A pure substance consists of only one element or one compound. A mixture consists of two or more different substances that are not chemically joined together.

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What process at the surface of the Earth is part of the formation of sedimentary rocks?

OLga [1]

Answer:

Clastic sedimentary rocks are made up of pieces (clasts) of pre-existing rocks. Pieces of rock are loosened by weathering, then transported to some basin or depression where sediment is trapped. If the sediment is buried deeply, it becomes compacted and cemented, forming sedimentary rock.

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3 years ago

When a solution of NaI reacts with a solution of AgNO3, what is the net ionic equation?

Leni [432]

Answer:

Ag⁺(aq) + I⁻(aq) → AgI(s)

Explanation:

Net ionic equation is a way to write a chemical equation in which you are listing only the species that are participating in the reaction.

In the reaction:

AgNO₃(aq) + NaI(aq) → AgI(s) + NaNO₃(aq).

The ionic equation is:

Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + I⁻(aq) → AgI(s) + Na⁺(aq) + NO₃⁻(aq).

Now, listing only the species that are participating in the reaction:

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3 years ago

The atomic number of an element is 71. is this element diamagnetic or paramagnetic?

Mrrafil [7]

The element with atomic number 71(lutetium) is paramagnetic

This is because all its electrons is not paired . lutetium has electron configuration of (Xe) 4f^14 5d^1 6S^2
Its 5d sub shell is not paired since the d sub shell can accommodate a total of 10 electrons hence lutetium is paramagnetic

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3 years ago

Read 2 more answers

A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of

zaharov [31]

Answer:

a) pH = 2.88

b) pH = 4.598

c) pH = 5.503

d) pH = 8.788

e) pH = 12.097

Explanation:

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.75 E-5 = [H3O+]*[CH3COO-] / [CH3COOH]

a) 0.0 mL KOH:

mass balance:

⇒ C CH3COOH = [CH3COOH] + [CH3COO-] = 0.100 M

charge balance:

⇒ [H3O+] = [CH3COO-]

⇒ 1.75 E-5 = [H3O+]²/(0.100 - [H3O+])

⇒ [H3O+]² + 1.75 E-5[H3O+] - 1.75 E-6 = 0

⇒ [H3O+] = 1.314 E.3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.88

b) 5.0 mL KOH:

CH3COOH + KOH ↔ CH3COONa + H2O

∴ C CH3COOH = ((0.025)(0.100) - (5 E-3)(0.200))/(0.025+5 E-3)

⇒ C CH3COOH = 0.05 M

∴ C KOH = ((5 E-3)(0,200))/(0.025+5 E-3) = 0.033 M

mass balance:

⇒ C CH3COOH + C KOH = [CH3COOH] + [CH3COO-] = 0.05 + 0.033 = 0.083 M

charge balance:

⇒ [H3O+] + [K+] = [CH3COO-]

⇒ [CH3COO-] = [H3O+] + 0.033

⇒ 1.75 E-5 = ([H3O+]*([H3O+] + 0.033))/(0.083 - ([H3O+] + 0.033))

⇒ 1.75 E-3 = ([H3O+]² + 0.033[H3O+])/(0.05 - [H3O+])

⇒ 8.75 E-7 - 1.75 E-5[H3O+] = [H3O+]² + 0.033[H3O+]

⇒ [H3O+]² +0.03302[H3O+] - 8.75 E-7 = 0

⇒ [H3O+] = 2.523 E-5 M

⇒ pH = 4.598

equivalent point:

(C*V)acid = (C*V)base

⇒ (0.100 M)*(0.025 L) = (0.200 M)( Vbase)

⇒ Vbase = 0.0125L = 12.5 mL

c) 10.0 mL KOH:

∴ C CH3COOH = 0.0143 M

∴ C KOH = 0.057 M

as in the previous point, starting from the mass and charge balances, we obtain:

⇒ [H3O+] = 3.1386 E-6 M

⇒ pH = 5.503

d) 12.5 mL KOH:

at the equivalence point, there is complete salt formation, then the pH is calculated through the salt:

CH3COO- + H2O ↔ CH3COOH - OH-

∴ Kw/Ka = 1 E-14/1.75 E-5 = 5.714 E-10 = [CH3COOH]*[OH-]/[CH3COO-]

∴ [CH3COO-] = (0.025)(0.100))/(0.025+0.0125) = 0.066 M

mass balance:

⇒ 0.066 = [CH3COOH] + [CH3COO-]..........(1)

charge balance:

⇒ [K+] = [OH-] + [CH3COO-] = 0.066 M.........(2)

∴ [K+] = C CH3COO- = 0.066 M

(1) = (2):

⇒ [OH-] = [CH3COOH].......(3)

⇒ 5.714 E-10 = [OH-]² / (0.066 - [OH-])

⇒ [OH-]² + 5.714 E-10[OH-] - 3.7712 E-11 = 0

⇒ [OH-] = 6.1408 e-6 m

⇒ pOH = 5.212

⇒ pH = 14 - pOH = 8.788

d) 15.0 mL KOH:

after the equivalence point there is salt and excess base (OH-); ph is calculated from excess base:

⇒ C KOH = ((0.015)(0.200) - (0.025)(0.100)) / (0.025 + 0.015) = 0.0125 M

⇒ [OH-] ≅ C KOH = 0.0125 M

⇒ pOH = 1.903

⇒ pH = 12.097

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3 years ago

What year will Aquarius be visible from Earth’s North Pole?

Minchanka [31]

Answer:

Aquarius beginning in the mid-3rd millennium. The north and south celestial poles are the two imaginary points in the sky where the Earth's ... In about 5,500 years, the pole will have moved near the position of the star ... The south celestial pole is visible only from the Southern Hemisphere.

Explanation:

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3 years ago