5. Which of the following atoms has the largest atomic

Enter the symbol of a sodium ion, Na+, followed by the formula of a sulfate ion, SO42−. Separate the ions with a comma only—spac

gladu [14]

Na⁺,SO₄²⁻ is the answer

<h3>Further explanation </h3>

An ion is an atom or molecule that has a net electrical charge. There are many ions, one of them are sodium ion and sulfate ion.

SO₄²⁻ or Sulfate is a naturally occurring polyatomic ion that consist of a central sulfur atom surrounded by four oxygen atoms with occured widely in everyday life. Sodium ions are important for regulation of blood and body fluids, transmission of nerve impulses, heart activity, and certain metabolic functions.

Whereas Na⁺ or Sodium is a chemical element with the symbol Na and atomic number 11. It is a soft, silvery-white, highly reactive metal. Sulfate ion is a very weak base. Therefore sulfate ion undergoes negligible hydrolysis in aqueous solution.

Enter the symbol of a sodium ion, Na⁺, followed by the formula of a sulfate ion, SO₄²⁻. Separate the ions with a comma only—spaces are not allowed. Express your answers as ions separated by a comma. Therefore the answer is: Na⁺,SO₄²⁻

Hope it helps!

<h3>Learn more</h3>

Learn more about sodium ion brainly.com/question/6839866
Learn more about sulfate ion brainly.com/question/2763823
Learn more about ions brainly.com/question/11852357

<h3>Answer details</h3>

Grade: 9

Subject: Chemistry

Chapter: Introduction to Mastering Chemistry

Keywords: sodium ion, sulfate ion, ions, Chemistry, symbol

6 0

2 years ago

Read 2 more answers

Consider the nuclear equation below. Superscript 124 subscript 56 B a right arrow superscript 124 subscript 55 upper C s plus qu

nadezda [96]

Superscript o subscript negative 1 e.

Explanation:

The nuclear reactions is of 2 types, one is nuclear fusion and the other one is nuclear fission.

Nuclear fusion is nothing but the combining of 2 nuclei with an emission of energy along with an electron, proton or beta particle.

Nuclear fission is the break down of a nucleus into 2 or more nuclei along with an electron, proton or beta particle.

And the reaction is,

₅₆B¹²⁴ ₅₅C¹²⁴ + ₋₁e⁰

So the blank was filled by means of a beta particle.

6 0

2 years ago

Read 2 more answers

What is the limiting reactant for the reaction below given that you start with 10.0 grams of Al (molar mass 26.98 g mol-1) and 1

mezya [45]

Answer:

Al

Explanation:

4 Al + 3 O₂ → 2 Al₂O₃

You need to figure out which one has the smaller mole ratio. Convert both substances from grams to moles.

(10.0 g Al)/(26.98 g/mol) = 0.3706 mol Al

(19.0 g O₂)/(32.00 g/mol) = 0.5938 mol O₂

Now, use the mole ratios of reactant to product to see which substance produces the least amount of product.

(0.3706 mol Al) × (2 mol Al₂O₃/4 mol Al) = 0.1853 mol Al₂O₃

(0.5938 mol O₂) × (2 mol Al₂O₃/3 mol O₂) = 0.3958 mol Al₂O₃

Since aluminum produces the least amount of product, this is the limiting reagent.

4 0

2 years ago

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N

Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

Now we have to calculate the dissociated concentration of $N_2O_4$ .

The balanced equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(aq)$

Initial conc. 1.0 M 0

At eqm. conc. (1.0-x) M (2x) M

As we are given,

The percent of dissociation of $N_2O_4$ = $\alpha$ = 28.0 %

So, the dissociate concentration of $N_2O_4$ = $C\alpha=1.0M\times \frac{28.0}{100}=0.28M$

The value of x = $C\alpha$ = 0.28 M

Now we have to calculate the concentration of $N_2O_4\text{ and }NO_2$ at equilibrium.

Concentration of $N_2O_4$ = 1.0 - x = 1.0 - 0.28 = 0.72 M

Concentration of $NO_2$ = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.56)^2}{0.72}=0.44$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.44

8 0

2 years ago

Give the product for the reaction of 1-butene with methanol in the presence of acid. The mechanism is the same as the mechanism

Nuetrik [128]

Answer:

The final result is <u>2-methoxybutane. </u>

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Explanation:

1-butene has a carbon-carbon double bound between C1 and C2.

When it will react with methanol, in the presence of an acid, the result will be an ether.

C4H8 + CH3OH → C5H12O

An acid-catalyzed ether synthesis from alkenes is limited by carbocation stability.

In the first step, the double bound will disappear. The C2 atom will be a C+ atom, this becaus it has only 3 bounds and not 4.

This C+ -atom will atract the O- atom to form an ether. The CH3 of methanol will bind on the C3 atom, this is the most stable position.

2-methoxybutane will be formed. It has a structural formula of C5H12O

1-methoxybutane will not be formed, because it's less stable.

1-butanol will be formed when water is added to 1-butene. The mechanism has the same principle but not the same product.

1-ethoxybutane and 2-ethoxybutane have a structural formula of C6H14O, this will not be the final result.

The final result is <u>2-methoxybutane. </u>

<u />

8 0

2 years ago