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3 years ago

What type of chemical bond forms between positive and negative ions? (1 point)?

Chemistry

2 answers:

Rainbow [258]3 years ago

8 0

Ionic is the correct answer

vlada-n [284]3 years ago

7 0

Ionic bond is formed between positive and negative ions.

Positive ion is known as cation (electropositive metal)

Negative ion is known as anion ( electronegative non-metal)

The bond formed between a metal (positive ion) and non-metal (negative ions) is known as Ionic bond.

Ionic form is formed when one or more electrons from the valence shell of an atom (electropositive metal ) are completely transferred to the valence shell of another atom (electronegative non-metal). Ionic bonding is the complete transfer of valence electron(s) between oppositely charged atoms.

For example: Bond between NaCl(sodium chloride) is an ionic bond in which Na (sodium ) is positive ion(metal) which have +1 charge and Cl (chlorine) is negative ion(non-metal) which have -1 charge and the bond formed between Na+ and Cl- is ionic bond.

$Na^{+}(positive ion) + Cl^{-}(negative ion)\rightarrow Ionic bond$

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Which statement correctly identifies the nuclide that is most likely to be unstable and describes why?

iVinArrow [24]

The answer would be B.

U-238 has a n to p ration of 1.6:1. 146 neutrons and 92 protons.

It is actually the most commonly used isotope is reactors.

C-14 is also a radioactive isotope with 8 neutrons and 6 protons.

The usual and ideal n to p ratio is 1:1 such as C-12 or Mg-24

8 0

3 years ago

How many moles of Calcium Oxide are needed to produce 4 moles of Calcium Hydroxide?

Tpy6a [65]

Taking into account the reaction stoichiometry, 2 moles of CaO are required to react with 2 moles of Ca(OH)₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CaO + H₂O → Ca(OH)₂

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

CaO: 1 mole
H₂O: 1 mole
Ca(OH)₂: 1 mole

<h3>Moles of CaO required</h3>

The following rule of three can be applied: If by stoichiometric reaction 1 mole of Ca(OH)₂ is produced by 1 mole of CaO, 2 moles of Ca(OH)₂ are produced by how many moles of CaO?

$moles of CaO=\frac{2 moles of Ca(OH)_{2}x1 mol of CaO }{1 mole of Ca(OH)_{2}}$

moles of CaO= 2 moles

Finally, 2 moles of CaO are required to react with 2 moles of Ca(OH)₂.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

5 0

2 years ago

An exothermic reaction is a chemical reaction that produces _____.

lana66690 [7]

I pretty sure the answer is a

8 0

3 years ago

The fizz produced when an Alka-Seltzer® tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO3)

cestrela7 [59]

Answer:

a. The limiting reactant is $NaHCO_{3}$

b. 0.73 g of carbon dioxide are formed.

c. The grams of excess reactant that do not participate in the reaction are 0333 g.

Explanation:

You know the following reaction:

$3NaHCO_{3} +H_{3} C_{6} H_{5} O_{7}$ ⇒ $3CO_{2} +3H_{2} O+Na_{3} C_{6} H_{5} O_{7}$

First, you determine the molar mass of each compound. For that you must take into account the atomic mass of each element:

Na: 23
H: 1
C: 12
O: 16

To determine the molar mass of each compound, you multiply the most atomic of each element present in the molecule by the sub-index that appears after each number, which indicates the present amount of each element in the compound:

$NaHCO_{3}$ :23+1+12+16*3=84 g/mol
$H_{3} C_{6} HO_{7}$ :1*3+12*6+1*5+16*7= 192 g/mol
$CO_{2}$ :12+16*2= 44 g/mol
$H_{2} O$ :1*2+16= 18 g/mol
$Na_{3} C_{6} H_{5} O_{7}$ : 23*3+12*6+1*5+16*7= 258 g/mol

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you know that 3 moles of $NaHCO_{3}$ react with 1 mole of $H_{3} C_{6} HO_{7}$ Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 192 g

You know that in a certain experiment you have 1.40 g of sodium bicarbonate and 1.40 g of citric acid. To determine the limiting reagent apply a rule of three simple as follows:

If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of sodium bicarbonate react with 1.4 grams of citric acid?

$grams of sodium bicarbonate= \frac{1.4 g*252 g}{192 g}$

grams of sodium bicarbonate= 1.8375 g

But to perform the experiment you have only 1.4 g of sodium bicarbonate. So <u><em>the limiting reagent is sodium bicarbonate</em></u>.

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first.

Now, by stoichiometry of the reaction, you know that 3 moles of $NaHCO_{3}$ react with 3 mole of $CO_{2}$ . Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 132 g

You make a simple rule of three: if 252 g of sodium bicarbonate form 132 g of carbon dioxide per stochetry, how many grams will form 1.4 g of sodium bicarbonate?

$grams of carbon dioxide =\frac{1.4 g * 132 g}{252 g}$

<u><em>grams of carbon dioxide= 0.73 g</em></u>

<u><em>Then, 0.73 g of carbon dioxide are formed.</em></u>

To know how much citric acid will react you apply a rule of three, taking into account as in the previous cases the stoichiometry of the reaction: If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of citric acid will they react with 1.4 g of sodium bicarbonate?

$grams of citric acid=\frac{1.4 g * 192 g}{252 g}$

grams of citric acid= 1.067 g

But you have 1.4 g of citric acid. That means that the grams you have minus the grams that react will be the grams that remain in excess and do not participate in the reaction:

grams of excess reactant=1.4 g - 1.067 g

grams of excess reactant=0.333 g

<em><u>So the grams of excess reactant that do not participate in the reaction are 0333 g.</u></em>

3 0

3 years ago

`Let's just consider the concept of stoichiometry without any confusing chemicals. Here is a generic equation: 2A + 6B ? 3C. If

lesya [120]

Answer:

Option D. 5.5

Explanation:

The equation is this:

2A + 6B ⇒ 3C

With the amounts that we were given, let's determine which is the <em>limting reactant</em>

2 A reacts with 6 B

4 A will react with ( 4 .6)/2 = 12B

I have 11 B, so the limiting is B

6 B react with 2 A

11 B will react with (11 .2 )/6 =3.66 A

I have 4 A, so A is the excess.

6 B produce 3 C

11 B will produce ( 11 .3)/6 = 5.5C

7 0

3 years ago