Answer: The molar concentration of sulfuric acid in the original sample is 1.943 M
Explanation:
To calculate the molarity of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Now to calculate the molarity of original solution:
Thus the molar concentration of sulfuric acid in the original sample is 1.943 M
Answer:
1. Chromium
2. Chlorine.
3. Chlorine.
4. Chromium.
5. 12 electrons.
Explanation:
Hello,
In this case, the given reaction with the appropriate oxidation states turns out:
In such a way, the oxidation half-reaction is written for chromium as the reducing agent so it is oxidized from +3 to +6, nonetheless, since there are two chromiums undergoing such change, 6 electrons are being transferred as shown below:
On the other hand, chlorine's reduction half-reaction as the oxidizing agent result from the transfer of 6 electrons as well from +1 to 0, nonetheless, there are 6 chlorines undergoing such change:
Therefore, there are 12 electrons that are being transferred, 6 for chromium and 6 for chlorine.
Best regards.
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
Answer:
Shifts the equilibrium to the left. reduces solubility.
Explanation:
- MgF2(s) ↔ Mg2+(aq) + 2F-(aq)
S S 2S
∴ Ksp = 6.4 E-9 = [ Mg2+ ] * [ F- ]² = S * (2S)²
⇒ 4S² * S = 6.4 E-9
⇒ 4S³ = 6.4 E-9
⇒ S³ = 1.6 E-9
⇒ S = 1.1696 E-3 M
- NaF(s) → Na+(aq) + F-(aq)
0.10M 0.10M 0.10M
- MgF2(s) ↔ Mg2+(aq) + 2F-(aq)
S' S' 2S' + 0.10
⇒ Ksp = 6.4 E-9 = (S')*(2S' + 0.10)²
If we compare the concentration (0.10 M) of the ion with Ksp ( 6.4 E-9 ); thne we can neglect S' as adding:
⇒ 6.4 E-9 = (S')*(0.10)² = 0.01S'
⇒ S' = 6.4 E-7 M
∴ % S' = ( 6.4 E-7 / 0.1 )*100 = 6.4 E-4% <<< 5%, we can make the assumption
We can observe that S >> S' ( 1.1696 E-3 M >> 6.4 E-7 M ), which shows that the solubility is reduced by the efect of the common ion from the salt, which causes the equilibrium to shift to the left, precipitating part of MgF2(s).
First one. Coefficients are numbers that balance the equation, just like if there is an equation in math where 1=2, you need to multiply 1 by 2 to make that equation true. That's a nice jingle you can remember.
