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3 years ago

Define cation and anion. In which case are electrons gained? Lost?

Chemistry

1 answer:

inn [45]3 years ago

5 0

Explanation:

A cation is a positively charged ion.

An anion is a negatively charged ion.

In a cation, electrons are usually lost
In an anion, electrons are usually gained.

In an atom, the balance between the number of protons and electrons determine whether they become an anion or cation.

Protons are positively charged particles.

Electrons are negatively charged

When an atom gains electrons, the number of electrons becomes more than that of protons. This leaves a net negative charge on the atom and it makes an anion.

When an atom loses electrons, the number of protons becomes more than that of electrons. This leaves a net positive charge on the atom and makes a cation.

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Which particle builds a static electric charge when it is transferred from one object to another?

murzikaleks [220]

The answer would be, "Electrons".

3 0

3 years ago

The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In t

const2013 [10]

Answer:

$\boxed{\text{47.4 g}}$

Explanation:

We are given the mass of two reactants, so this is a limiting reactant problem.

We know that we will need mases, moles, and molar masses, so, let's assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r: 17.03 32.00 18.02

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

m/g: 70.1 70.1

Step 1. Calculate the moles of each reactant

$\text{Moles of CO } = \text{70.1 g} \times \dfrac{\text{1 mol}}{\text{17.03 g}} = \text{4.116 mol}\\\\\text{Moles of H$_{2}$O} = \text{70.1 g} \times \dfrac{\text{1 mol}}{\text{32.00 g}} = \text{2.191 mol}$

Step 2. Identify the limiting reactant

Calculate the moles of H₂O we can obtain from each reactant.

From NH₃:

The molar ratio of H₂O:NH₃ is 6:4.

$\text{Moles of H$_{2}$O} = \text{4.116 mol NH$_{3}$} \times \dfrac{\text{6 mol H$_{2}$O}}{\text{4 mol NH$_{3}$}} = \text{6.174 mol H$_{2}$O}$

From O₂:

The molar ratio of H₂O:O₂ is 6:5.

$\text{Moles of H$_{2}$O} = \text{2.191 mol O$_{2}$} \times \dfrac{\text{6 mol H$_{2}$O}}{\text{5 mol O$_{2}$}} = \text{2.629 mol H$_{2}$O}$

O₂ is the limiting reactant because it gives the smaller amount of H₂O.

Step 3. Calculate the theoretical yield.

$\text{Theor. yield } = \text{2.629 mol H$_{2}$O}\times \dfrac{\text{18.02 g H$_{2}$O}}{\text{1 mol H$_{2}$O}} = \textbf{47.4 g H$_{2}$O}\\\\\text{The maximum yield of H$_{2}$O is }\boxed{\textbf{47.4 g}}$

6 0

3 years ago

The two naturally occurring isotopes of antimony, 121Sb (57.21 percent) and 123Sb (42.79 percent), have masses of 120.904 and 12

Alex

Answer:

The correct answer is option c.

Explanation:

Formula used to determine an average atomic mass :

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

Mass of isotope Sb-121 = 120.904 amu

Fractional abundance of Sb-121 = 57.21% = 0.5721

Mass of isotope Sb-123 = 122.904 amu

Fractional abundance of Sb-123 = 42.79% = 0.4279

Average atomic mass of Sb:

$120.904 amu\times 0.5721+ 122.904 amu\times 0.4279=121.7598 amu \approx 121.76 amu$

7 0

3 years ago

Calculate the percentage composition of Mg3 (Po4)2

lutik1710 [3]

Mg3(PO4)2 - the molar mass would be 262g/mol, which is 100%

Atomic mass of Mg is 24, since we have 3Mg we multiply by 3 and get a mass of 72

262 : 100% = 72 : x%

x = 72*100 / 262

x = 27.5%

And do that for every element — get the molar mass of P and multiply by 2, use a ratio, and get the molar mass of O and multiply by 8 and use ratios :)

7 0

3 years ago

How many grams of F are in 12.56 g of SF6? h.

natali 33 [55]

Answer:

9.80 g

Explanation:

The molecular mass of the atoms mentioned in the question is as follows -

S = 32 g / mol

F = 19 g / mol

The molecular mass of the compound , SF₆ = 32 + ( 6 * 19 ) = 146 g / mol

The mass of 6 F = 6 * 19 = 114 g /mol .

The percentage of F in the compound =

mass of 6 F / total mass of the compound * 100

Hence ,

The percentage of F in the compound = 114 g /mol / 146 g / mol * 100

78.08 %

Hence , from the question ,

In 12.56 g of the compound ,

The grams of F = 0.7808 * 12.56 = 9.80 g

4 0

3 years ago