The pH of an aqueous solution that has a concentration of 0.35 M NaF and pKa for HF = 3.14 is 3.6.
<h3>How to calculate pH?</h3>
The pH of a solution refers to the degree of acidity or alkalinity of the solution. It can be calculated using the Henderson-Hasselbalch Equation as follows:
pH = pka + log ([A-]/[HA])
Where;
- A- = conjugate base
- HA = weak acid
pH = pKa + log([F-]/[HF])
pH = 3.14 + log(1/0.35)
pH = 3.14 + 0.4559 = 3.595
Therefore, the pH of an aqueous solution that has a concentration of 0.35 M NaF and pKa for HF = 3.14 is 3.6.
Learn more about pH at: brainly.com/question/15289741
<span>the right answer is .C )<span>Matthew was right because RNA is single-stranded while DNA is double-stranded.</span></span>
Answer:
V₂ = 918.1 cm³
Explanation:
Given data:
Initial volume = 640 cm³
Initial temperature = 100°C (100+273 = 373 K)
Initial pressure = 1490 mmHg (1490 /760 = 1.96 atm)
Final volume = ?
Final temperature = 273 K
Final pressure = 1 atm
Solution:
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
now we will put the values in formula.
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 1.96 atm × 640 cm³ × 273 K / 373 K × 1 atm
V₂ = 342451.2 atm .cm³ . K / 373 K. atm
V₂ = 918.1 cm³
Answer:
scientists are uncertain of how long the resources will last.
Explanation:
It is hard to estimate how long certain natural resources will last because they are dependent upon the rate of consumption and other factors. There are many fields of studies going into alternative energy. Since the technology needed for some of these ideas will take time to be developed, alternative sources of energy must be found before the resources we already have run out.
The balanced chemical reaction is expressed as:
M + F2 = MF2
To determine the moles of the element fluorine present in the product, we need to determine the moles of the product formed from the reaction and relate this value to the ratio of the elements in MF2. We do as follows:
moles MF2 produced = 0.600 mol M ( 1 mol MF2 / 1 mol M ) = 0.600 mol MF2
molar mass MF2 = 46.8 g MF2 / 0.6 mol MF2 = 78 g/mol
moles MF2 = 46.8 g ( 1 mol / 78 g ) = 0.6 mol
moles F = 0.6 mol MF2 ( 2 mol F / 1 mol MF2 ) = 1.2 moles F
