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3 years ago

Which compound is insoluble in water? (1) KOH (3) Na3PO4 (2) NH4Cl (4) PbSO4

Chemistry

2 answers:

VLD [36.1K]3 years ago

8 0

Ans: 4) PbSO4

Different compounds are soluble in water to a different extent, some of them are completely insoluble in water. Solubility rules help in identifying the water soluble and insoluble compounds.

Based on the rules:

1) Salts of group IA (Li, Na, K, Rb and Cs) are soluble in water

2) Salts containing NH4+ ions are water soluble

3) All phosphates (PO4³⁻) are insoluble except those containing group IA elements and (NH4)₃PO4

4) All sulfates, SO4²⁻ salts are generally soluble with the exception of Ca2+, Sr2+, Ba2+, Hg₂²⁺, Hg2+, Pb2+ and Ag+

Therefore, among the given examples:

1) KOH contains group 1A (K) therefore soluble

2) Na3PO4 contains group IA element (Na), therefore soluble

3) NH4Cl contains NH4+ ion, therefore soluble

4) Sulfate of lead (Pb) is insoluble

Veseljchak [2.6K]3 years ago

3 0

The answer is (4) PbSO₄.

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2Al+6HBr -> 2AlBr3+3H2. When 3.22 moles of Al reacts with 4.96 moles of HBr, how many moles of H2 are formed?

Cerrena [4.2K]

2 Al+6 HBr = 2 AlBr₃ + 3 H₂

2 moles Al --------- 6 moles HBr ----------- 3 moles H₂
3.22 moles Al ------ 4.96 moles HBr ----- ( moles H₂ )

moles H₂ = 4.96 x 3 / 6

moles H₂ = 14.88 / 6

= 2.48 moles of H₂

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3 years ago

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Two gases, A and B, are at equilibrium in a sealed cylinder. Individually, gas A is colorless, while gas B is dark colored. The

Dmitriy789 [7]

Answer:

(a) color

(b) endothermic

(c)

b The ΔH value would have the same magnitude value but opposite sign.

c The K expression would be inverted.

Explanation:

Let's consider the following reaction at equilibrium.

A(g) ⇄ 2 B(g)

colorless dark colored

(a) The cylinder should appear (color or colorless)

At equilibrium, there is a mixture of A and B, so the cylinder should appear colored.

(b) When the system is cooled, the cylinder's appearance becomes very light colored. Therefore, the reaction must be (endothermic or exothermic)

According to Le Chatelier's Principle when a perturbation is made to a system at equilibrium it will react to counteract such effect. When the system is cooled, it will tend to increase the temperature by releasing heat. In this case, the reaction is endothermic so when the reverse reaction is favored, colorless A is favored as well.

Suppose the reaction equation were written as follows: 2 B(g) ⇄ A(g)

(c) Which of these statements would then be true?

a The value of K would not change. FALSE. The new K would be the inverse of the direct K.

b The ΔH value would have the same magnitude value but opposite sign. TRUE. This is stated by Lavoisier-Laplace Law.

c The K expression would be inverted. TRUE. What was product before now is reactant and vice-versa.

d The color of the cylinder would be darker. FALSE. Changing the way the reaction is expressed has no effect on the equilibrium.

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4 years ago

50 POINTS PLUS BRAINLIEST NEED HELP ASAP Please solve these 5 questions

MariettaO [177]

Answer:

Argon 7.89693x10^25 atoms

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3 years ago

This graph shows two curves pertaining to a hydrogen s orbital.

fgiga [73]

Answer 1) : According to the complete question attached in the answer,

The radial wave function which is denoted by $R_{nl}(r)$ shown with orange color crosses through zero point. Also, At the the radial nodes, which are spherical shells to some radial distance away from the nucleus there no electron are found.

Also, the radial probability distribution curve denoted as $R^{2}_{nl}(r)$ shown in blue color is observed to touch zero, and shows the place of radial node.

Therefore, the total number of nodes will include both the kinds which has radial and angular nodes which will be represented by 'n'.

It is observed that for any atomic orbital, the total number of nodes will be n-1 .

Considering the s orbital of the hydrogen, which has zero angular momentum (l); (l=0), as it has zero angular nodes.

Hence, there will be only radial nodes, which is

(n−1 = total number of radial nodes in s orbitals)

According to the image, there are 4 radial nodes shown, so n = 5 (as n-1 = 4; therefore, n = 5)

This represents the 5s orbital.

Answer 2) The radial nodes are observed in I'm seeing radial nodes at

$1.9a_{0}$ , $6.4a_{0}$ , $13.9a_{0}$ and $27.0a_{0}$ .

where $a_{0}$ represents the hydorgen bohr atomic radius = 0.0529177 nm

Explanation : It is quite easy to observe the given graph and find out the approximate values of the radial nodes, it does not requires any equation to be solved. Equation can be used to find the radial nodes if it was supplied along with the question. Although by mere speculation one can find out the answer.

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The correct answer is B) Double displacement.

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It helps to remember that when there is a double replacement there are 4 elements/molecules in the reactants and 4 in the products that get rearranged.

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3 years ago

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