Answer:
Change in height of the object
Explanation:
Change in height of the object
Since ,
We know potential energy can be written as mgh where
P.E. = mgh
M = mass ,
g = gravity constant
h = height
Kinetic energy can be written as 0.5 mv²
So , the Potential energy = kinetic energy
Then ,
velocity of object = √2 gh
Since g is a constant ,
Hence ,
Height of object will determine the velocity.
Answer: -
Many drugs are sold as their hydrochloric salts (RNH₃⁺Cl⁻), formed by reaction of an amine (RNH₂) with HCl.
It is done because amines are generally liquids. But their hydrochloric salts are solid. A solid drug is always more preferable for drug companies as their handling and packaging are easier.
Acebutolol consists of one amide functionality as well as a secondary amine functionality.
When HCl is added, the lone pairs of the nitrogen of the secondary amine attacks it, leading to the formation of it's hydrochloric salts.
Answer:
q
Explanation:
The product and balanced net ionic equations for the following reactions are SnCl₂ + 2KMnO₄ ⇒ 2 KCl + Sn(MnO₄)₂.
Ionic equations are those equations that happened in an aqueous solution. The chemical equation is expressed in an electrolyte solution is expressed and dissociates ions.
In these reactions, each element or ion is dissociated into differently charged ions in a solution. Each of the ions is shown with different charges.
SnCl₂ + 2KMnO₄ ⇒ 2 KCl + Sn(MnO₄)₂. In the reaction, the tin chloride, and potassium magnesium oxide. It dissociates into charged ions, like potassium chloride and tin magnesium oxide. The chlorine will acquire a negative charge and magnesium oxide get a positive charge.
Thus, the net ionic equation is SnCl₂ + 2KMnO₄ ⇒ 2 KCl + Sn(MnO₄)₂.
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Answer:
The answer to your question is:
Explanation:
Reaction
SnCl₂ + 2KMnO₄ ⇒ 2 KCl + Sn(MnO₄)₂
1 ---- Sn ---- 1
2 ---- K ----- 2
2 ---- Mn ---- 2
8 ---- O ---- 8
2 ---- Cl ---- 2
Answer:
Zeros located at the end of significant figures are significant.
Explanation:
Hope it will help :)
Answer:
Well atomic number 17 is Chlorine, which is most commonly found as a gas, and is period 7.
Explanation:
elements found on period 7 are some of the most unstable elements.
b. 3.35
c. 2.41
d. 1.48
e. 7.00
Answer:
b. 3.35
Explanation:
To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.
pH = pKa + log ([salt]/[acid]) (Eq. 01)
Where
pKa = -log(Ka) (Eq. 02)
[salt] = Molar concentration of salt produced as a result of titration
[acid] = Molar concentration of acid left in the solution after titration
Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:
HNO2 + KOH ⇆ H2O + KNO2 (Eq. 03)
This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.
Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles
Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles
As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.
Therefore
Amount of salt produced i.e [salt] = 0.0025 moles (Eq. 04)
Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)
Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:
pH= -log(4.5x10 -4) + log (0.0025/0.0025)
Solving above we get
pH = 3.35
The pH value in the titration flask after 25.00 mL of the 0.10 M KOH solution is added to 50.00 mL of 0.10 M HNO2 solution is 3.35.
The subject of this question is titration, which is a method used in chemistry to measure the concentration of an unknown solution. Given 50.00 mL of 0.10 M HNO2 (nitrous acid, Ka = 4.5 × 10-4), titrated with 0.10 M KOH (potassium hydroxide), we need to calculate the pH after 25.00 mL of the KOH solution is added.
First, we need to find the moles of the HNO2 and the KOH. Moles equals Molarity times Volume. So, for HNO2, it is 0.10 M * 0.050 L which equals 0.005 moles. For KOH, it is 0.10 M * 0.025 L which equals 0.0025 moles.
Then, subtract the moles of OH- from the moles of HNO2 to determine the concentration of HNO2 left, which is 0.005 moles - 0.0025 moles = 0.0025 moles. Divide this by the total volume of the solution (50.00 mL + 25.00 mL = 75.00 mL or 0.075 L to determine the new concentration of HNO2, 0.0025 moles / 0.075 L = 0.033 M. Then use the given Ka value with the equation [H+] = sqrt(Ka * [HNO2]) to get [H+].
To find acids' pH, we use the formula pH = -log[H+]. Use the calculated [H+] to find the pH.
Upon performing these calculations, the resulting pH value should be approximately 3.35 after 25.00 mL of the KOH solution is added, so the answer is (b) 3.35.
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