You’ll get marked as brainlist I think because that’s what happened last time (:

1. Which liquid sample is a pure substance?

IRISSAK [1]

The whole Activity , poem and paragraph is missing in the question.

Answer:

(1) Liquid A

(2) Solid A

Explanation:

Using this part of the given poem

Substances and mixtures behave differently,

During boiling and melting most especially

Boiling point of substance is fixed while mixture is not

Substance melts completely but mixture does not

The boiling point of the Pure substance remain fixed after reaching its boiling point this is shown by Liquid A

Solid A is melting completely so Solid A is a pure substance.

6 0

3 years ago

Which soil would be the best for growing plants?

Leokris [45]

Honestly, its based on your preference but the best recommended one would be loam. I use loam too, and it is really good for growing the plants.

:)

7 0

3 years ago

Which one of the following solutions will have the greatest concentration of hydroxide ions?

jek_recluse [69]

Answer:

e = 0.250 M calcium hydroxide

it will produce 0.5 M [OH⁻].

Explanation:

A = 0.100 M HCl

HCl is an acid it would produce negligible amount of OH⁻.

B = 0.100 M magnesium hydroxide

Mg(OH)₂ + H₂O → Mg²⁺ + 2OH⁻

The ration of Mg(OH)₂ and OH⁻ is 1:2.

[OH⁻] = 2× 0.100 = 0.200 M

C =0.100 M ammonia

NH₃ + H₂O → NH₄OH → NH₄⁺ + OH⁻

The Kb expression will be written as,

Kb = [NH₄⁺] [OH⁻] / [NH₃]

Kb = 1.8 ×10 ⁻⁵

Now we will put the values:

1.8 ×10 ⁻⁵ = [NH₄⁺] [OH⁻] / 0.100

[NH₄⁺] [OH⁻] = x²

1.8 ×10 ⁻⁵ = x² / 0.100

x² = 1.8 ×10 ⁻⁵ × 0.100

x² = 1.8 × 10 ⁻⁶

x = 1.34 × 10 ⁻³

[OH⁻] = 1.34 × 10 ⁻³ M

d = 0.300 M rubidium hydroxide

RbOH + H₂O → Rb⁺ + OH⁻

The ratio of RbOH and OH⁻ is 1:1.

[OH⁻] = 0.300 M

e = 0.250 M calcium hydroxide

Ca(OH)₂ + H₂O → Ca²⁺ + 2OH⁻

The ration of Ca(OH)₂ and OH⁻ is 1:2.

[OH⁻] = 2× 0.250 = 0.5 M

7 0

3 years ago

How many nonmetal elements are on the periodic table of elements

vladimir1956 [14]

There are 17 nonmetals on the periodic table of elements.

3 0

3 years ago

Read 2 more answers

is the general formula of a certain hydrate. When 256.3 g of the compound is heated to drive off the water, 214.2 g of anhydrous

Marizza181 [45]

Answer:

The general formula of the hydrate is Caa Seb Oc. nH2O. Based on the given information, the weight of the hydrated compound is 256.3 grams, the weight of the anhydrous compound is 214.2 grams.

Therefore, the weight of water evaporated is 256.3 g - 214.2 g = 42.1 grams

The molecular weight of water is 18 gram per mole. So, the number of moles of water will be,

Moles of water = weight of water/molecular weight

= 42.1 grams / 18 = 2.3

The given composition of calcium is 21.90 %. So, the concentration of calcium in anhydrous compound is,

= 214.2 * 0.2190 = 46.91 grams

The given composition of Se is 43.14 %. So, the concentration of selenium in anhydrous compound is,

= 214.2 * 0.4314 = 92.40 grams

The given composition of oxygen is 34.97%, So, the concentration of oxygen in anhydrous compound is,

= 214.2 * 0.3497 = 74.91 grams

The molecular weight of Ca is 40.078, the obtained concentration is 46.91 grams, stoichiometry will be, 46.91/40.078 = 1.17

The molecular weight of Se is 78.96, the obtained concentration is 92.40, stoichiometry will be,

92.40/78.96 = 1.17

The molecular weight of Oxygen is 15.999, the concentration obtained is 74.91, the stoichiometry will be,

74.91/15.999 = 4.68.

Thus, the formula becomes, Ca1.17. Se1.1e O4.68. 2.3H2O, the closest actual component is CaSeO4.2H2O

6 0

4 years ago