Answer:
a. 3; b. 5; c. 10; d. 12
Explanation:
pH is defined as the negative log of the hydronium concentration:
pH = -log[H₃O⁺] (hydronium concentration)
For problems a. and b., HCl and HNO₃ are strong acids. This means that all of the HCl and HNO₃ would ionize, producing hydronium (H₃O⁺) and the conjugate bases Cl⁻ and NO₃⁻ respectively. Further, since all of the strong acid ionizes, 1 x 10⁻³ M H₃O⁺ would be produced for a., and 1.0 x 10⁻⁵ M H₃O⁺ for b. Plugging in your calculator -log[1 x 10⁻³] and -log[1.0 x 10⁻⁵] would equal 3 and 5, respectively.
For problems c. and d. we are given a strong base rather than acid. In this case, we can calculate the pOH:
pOH = -log[OH⁻] (hydroxide concentration)
Strong bases similarly ionize to completion, producing [OH⁻] in the process; 1 x 10⁻⁴ M OH⁻ will be produced for c., and 1.0 x 10⁻² M OH⁻ produced for d. Taking the negative log of the hydroxide concentrations would yield a pOH of 4 for c. and a pOH of 2 for d.
Finally, to find the pH of c. and d., we can take the pOH and subtract it from 14, giving us 10 for c. and 12 for d.
(Subtracting from 14 is assuming we are at 25°C; 14, the sum of pH and pOH, changes at different temperatures.)
Answer:
The pH of the solution is 5.31.
Explanation:
Let "
is the dissociation of weak acid - HCN.
The dissociation reaction of HCN is as follows.
Initial C 0 0
Equilibrium c(1- ) c c
Dissociation constant = 
In this case weak acids is very small so, (1- ) is taken as 1.
From the given the concentration = 0.050 M
Substitute the given value.
![[H_{3}O^{+}]=c\alpha](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%5D%3Dc%5Calpha)
![[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%5D%3D0.05%5Ctimes%209.8%5Ctimes%2010%5E%7B-4%7D%3D%204.9%5Ctimes10%5E%7B-6%7D)
![pH= -log[H_{3}O^{+}]](https://tex.z-dn.net/?f=pH%3D%20-log%5BH_%7B3%7DO%5E%7B%2B%7D%5D)
![=-log[4.9\times10^{-6}]](https://tex.z-dn.net/?f=%3D-log%5B4.9%5Ctimes10%5E%7B-6%7D%5D)
Therefore, The pH of the solution is 5.31.
Answer:
Mass of SO₄⁻² = 0.123 g.
Mass percentage of SO₄⁻² = 41.2%
Mass of Na₂SO₄ = 0.0773 g
Mass of K₂SO₄ = 0.1277 g
Explanation:
Here we have
We place Na₂SO₄ = X and
K₂SO₄ = Y
Therefore
X +Y = 0.205 .........(1)
Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have
Amount of BaSO₄ from Na₂SO₄ is therefore;
Amount of BaSO₄ from K₂SO₄ is;
Molar mass of
BaSO₄ = 233.38 g/mol
Na₂SO₄ = 142.04 g/mol
K₂SO₄ = 174.259 g/mol
= 
= 1.643·X
= 
= 1.339·Y
Therefore, we have
1.643·X + 1.339·Y = 0.298 g.....(2)
Solving equations (1) and (2) gives
The mass of SO₄⁻² in the sample is given by
Mass of sample = 0.298
Molar mass of BaSO₄ = 233.38 g/mol
Mass of Ba = 137.327 g/mol
∴ Mass of SO₄ = 233.38 g - 137.327 g = 96.05 g
Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412
Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.
Percentage mass of SO₄⁻² = 41.2%
Solving equations (1) and (2) gives
X = 0.0773 g and Y = 0.1277 g.
Answer:
Te
K
Explanation:
Main group elements are elements designated and located on the s-block and p-block on the periodic table. The core group or prominent elements are the elements that make up those groups. These groupings comprise the most naturally rich and abundant elements. The main group elements are those whose characteristics are more predicted as a function of their periodic table location. From the given options, only Te and K are elements of the main-group.
Radio waves can harm people, but due to the wavelength of the X-Rays which are extremely dangerous to humans, and the wavelength of Radio waves (which are calm compared to the way X-Rays look) compared side by side, it would take extensive exposure to the Radio waves to cause disease, cancer, or even death. Hope I helped!
