Molality of C2H5OH is 1.1.27m.
What is Molality?
Molality is no.of moles present in One Kg solution .
Molality is represented by m
m= no.of moles/ weigt of solution in kg
Given is Molarity= 51.30M
molar mass of C2H5OH is 46g / mol , Density =0.9349g/ ml
Density=m/V
V = 107 ml
Molarity= no of moles/ Volume
51.5×46×107= x × 1000
weight of C2H5OH = 27.7g
molality = 27.7/1000××46
m= 1.27m
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Answer:
hi there !!
B) Water evaporates into steam.
this is correct because only physical state changes, steam can be cooled down to get water again.
In buffer solution there is an equilibrium between the acid HA and its conjugate base A⁻: HA(aq) ⇌ H⁺(aq) + A⁻(aq).
When acid (H⁺ ions) is added to the buffer solution, the equilibrium is shifted to the left, because conjugate base (A⁻) reacts with hydrogen cations from added acid, according to Le Chatelier's principle: H⁺(aq) + A⁻(aq) ⇄ HA(aq). So, the conjugate base (A⁻) consumes some hydrogen cations and pH is not decreasing (less H⁺ ions, higher pH of solution).
A buffer can be defined as a substance that prevents the pH of a solution from changing by either releasing or absorbing H⁺ in a solution.
Buffer is a solution that can resist pH change upon the addition of an acidic or basic components and it is able to neutralize small amounts of added acid or base, pH of the solution is relatively stable
Answer:
36.66%
Explanation:
Step 1: Given data
- Mass of the sample: 2.875 g
Step 2: Calculate the mass of salt
The mass of the sample is equal to the sum of the masses of the components.
m(sample) = m(iron) + m(sand) + m(salt)
m(salt) = m(sample) - m(iron) - m(sand)
m(salt) = 2.875 g - 0.660 g - 1.161 g
m(salt) = 1.054 g
Step 3: Calculate the percent of salt in the sample
We will use the following expression.
%(salt) = m(salt) / m(sample) × 100%
%(salt) = 1.054 g / 2.875 g × 100% = 36.66%
