Answer: The weight/weight % or percent by mass of the solute is 5.41 %.
Explanation:
Mass of the sodium sulfate,w = 9.74 g
Volume of the water = 165 mL
Density of the water = 1 g/mL

Mass of the water =
Mass of the solution, W:
Mass of solute + Mass of solvent =9.47 g + 165 g=174.47 g

The weight/weight % or percent by mass of the solute is 5.41 %.
5 examples of gases found in the normal home environment include; oxygen (air), nitrogen (most abundant element found in the air), carbon (air), a slight trace of argon and finally, hydrogen. these are- Nitrogen, oxygen, carbondioxide, carbonmonoxide and SO2.
Answer:
0.6749 M is the concentration of B after 50 minutes.
Explanation:
A → B
Half life of the reaction = 
Rate constant of the reaction = k
For first order reaction, half life and half life are related by:


Initial concentration of A = ![[A]_o=0.900 M](https://tex.z-dn.net/?f=%5BA%5D_o%3D0.900%20M)
Final concentration of A after 50 minutes = ![[A]=?](https://tex.z-dn.net/?f=%5BA%5D%3D%3F)
t = 50 minute
![[A]=[A]_o\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_o%5Ctimes%20e%5E%7B-kt%7D)
![[A]=0.900 M\times e^{-0.02772 min^{-1}\times 50 minutes}](https://tex.z-dn.net/?f=%5BA%5D%3D0.900%20M%5Ctimes%20e%5E%7B-0.02772%20min%5E%7B-1%7D%5Ctimes%2050%20minutes%7D)
[A] = 0.2251 M
The concentration of A after 50 minutes = 0.2251 M
The concentration of B after 50 minutes = 0.900 M - 0.2251 M = 0.6749 M
0.6749 M is the concentration of B after 50 minutes.
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Answer:
5.17.
Explanation:
<em>∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
</em>
[OH⁻] = 1.5 x 10⁻⁹ M.
∴ [H₃O⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/(1.5 x 10⁻⁹ M) = 6.66 × 10⁻⁶ M.
∵ pH = - log[H₃O⁺]
<em>∴ pH = - log(6.66 × 10⁻⁶ M) = 5.17.
</em>