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4 years ago

A bond formed when atoms transfer elentron is a bond

1 answer:

5 0

Ionic bonds involve a cation and an anion. The bond is formed when an atom, typically a metal, loses an electron or electrons, and becomes a positive ion, or cation. Another atom, typically a non-metal, is able to acquire the electron(s) to become a negative ion, or anion.

One example of an ionic bond is the formation of sodium fluoride, NaF, from a sodium atom and a fluorine atom. In this reaction, the sodium atom loses its single valence electron to the fluorine atom, which has just enough space to accept it. The ions produced are oppositely charged and are attracted to one another due to electrostatic forces.

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Each of the diagrams below shows an electrically neutral atom. fill in the missing number of protons (n) neutrons (and), electro

vova2212 [387]

What is the diagram?

7 0

3 years ago

What volume (in L) of a 1.25 M

sleet_krkn [62]

Answer:

V1= 0.305L

Explanation:

To find the initial volume of 1.25M potassium fluoride needed to make tge dilution specified in the question, we can use: C1 × V1 = C2 × V2

Since the question wants the volume in litres, convert 455 mL to L

455/ 1000

= 0.455 L

Now make the substitution

1.25 × V1 = 0.838 × 0.455

Rearrange to make V1 the subject

V1=

$\frac{0.838 \times 0.455}{1.25 } = 0.305$

4 0

4 years ago

State the oxidation number assigned to each bold element in the formula: NH4+1 a 3 b -3 c -1 d 6

Leya [2.2K]

The fomula is NH4 (1+)

There are only two elements N and H.

As per oxidation state rules, the most electronegative element will have a negative oxidation state and the other element will have a positive oxidation state.

N is more electronative than H, so H will have a positive oxidation state and nitrogen will have a negative oxidation state.

You can also use the rule that states the hydrogen mostly has 1+ oxidation state,except when it is bonded to metals.

In conclusion the oxidation state of H in NH4 (1+) is 1+.

Now you must know that the sum of the oxidations states equals the charge of the ion, which in this case is 1+.

That implies that 4* (1+) + x = 1+

=> x = (1+) - 4(+) = 3-

Answer: the oxidation state of N is 3-, that is the option b.

8 0

4 years ago

A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal

goblinko [34]

Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂:

$1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC$

All of the hydrogens atoms in the compound ended up becoming H₂O, so the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O:

$0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH$

Because the compound is composed only by C, H and O, the mass of O in the compound can be calculated by substraction:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C

mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H

mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

8 0

3 years ago

What mass, in grams, of CO2 and H2O is formed from 2.55 mol of propane?

oksian1 [2.3K]

Answer:

336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.

Explanation:

In this case, the balanced reaction is:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:

C₃H₈: 1 mole
O₂: 5 moles
CO₂: 3 moles
H₂O: 4 moles

Being the molar mass of each compound:

C₃H₈: 44 g/mole
O₂: 16 g/mole
CO₂: 44 g/mole
H₂O: 18 g/mole

Then, by stoichiometry, the following quantities of mass participate in the reaction:

C₃H₈: 1 mole* 44 g/mole= 44 grams
O₂: 5 moles* 16 g/mole= 80 grams
CO₂: 3 moles* 44 g/mole= 132 grams
H₂O: 4 moles* 18 g/mole= 72 grams

So you can apply the following rules of three:

If by stoichiometry 1 mole of C₃H₈ forms 132 grams of CO₂, 2.55 moles of C₃H₈ how much mass of CO₂ will it form?

$mass of CO_{2} =\frac{2.55 moles of C_{3} H_{8}*132 gramsof CO_{2} }{ 1 mole of C_{3} H_{8}}$

mass of CO₂= 336.6 grams

If by stoichiometry 1 mole of C₃H₈ forms 72 grams of H₂O, 2.55 moles of C₃H₈ how much mass of H₂O will it form?

$mass of H_{2}O =\frac{2.55 moles of C_{3} H_{8}*72 gramsof H_{2}O }{ 1 mole of C_{3} H_{8}}$

mass of H₂O= 183.6 grams

336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.

3 0

3 years ago