Science experiments usually include an independent variable, dependent variable, and control. ... Science experiments also include something called constants. A constant is the part that doesn't change during the experiment.
Answer:
ok but this number is for school another one number soon tell you
Answer:
the pH of HCOOH solution is 2.33
Explanation:
The ionization equation for the given acid is written as:

Let's say the initial concentration of the acid is c and the change in concentration x.
Then, equilibrium concentration of acid = (c-x)
and the equilibrium concentration for each of the product would be x
Equilibrium expression for the above equation would be:
![\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}](https://tex.z-dn.net/?f=%5CKa%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BHCOO%5E-%5D%7D%7B%5BHCOOH%5D%7D)

From given info, equilibrium concentration of the acid is 0.12
So, (c-x) = 0.12
hence,

Let's solve this for x. Multiply both sides by 0.12

taking square root to both sides:

Now, we have got the concentration of ![[H^+] .](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20.)
![[H^+] = 0.00465 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20%3D%200.00465%20M)
We know that, ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
pH = -log(0.00465)
pH = 2.33
Hence, the pH of HCOOH solution is 2.33.
Answer:
True
Explanation:
The physical and chemical properties of a substance depend on the nature of intermolecular forces between its molecules. For instance, water has a high boiling point because of hydrogen bonding between water molecules. Liquid water is denser than ice because of the difference in the nature of hydrogen bonding in liquid water and ice.
Answer:
The initial temperature was 58.4°C
Explanation:
Given the following data:
initial volume = V₁ = 380 mL = 0.38 L
final volume = V₂ = 250 mL = 0.25 L
final temperature = T₂ = -55°C = 218 K
According to Charles's law, the volume of a gas is <em>directly proportional to the temperature</em> (in Kelvin). The mathematical expression is:
V₁/T₁= V₂/T₂
So, we calculate the initial temperature (V₁) as follows:
T₁ = T₂/V₂ x V₁ = 218 K/(0.25 L) x 0.38 L = 331.36 K ≅ 331.4 K
Finally, we convert the initial temperature from K to °C:
T₁= 331.4 K - 273 = 58.4°C