Answer:
Explanation:
[ H⁺] = 3.5 x 10⁻⁶ M .
[ H⁺] [ OH⁻] = 10⁻¹⁴
[ OH⁻] x 3.5 x 10⁻⁶ = 10⁻¹⁴
[ OH⁻] = 2.857 x 10⁻⁹ M .
Answer:
Clay
Explanation:
Since Clay has a very small particle size it compacts or has a high surface area which slows the path of water, holding it for longer periods compared to other soil components.
Answer:
False
Step-by-step explanation:
The trough-to-trough distance is the <em>wavelength</em> (λ) of a wave.
The <em>amplitude</em> (a) of a wave is half the peak-to-trough distance. It is also the distance from the trough to the rest position.
Here, we are required to determine the pressure of the sample of gas trapped in the open-tube mercury manometer shown below if Atmospheric pressure is 767 mmHg and pressure head, h = 8.9 cmHg.
- The pressure of the sample of gas trapped in the open-tube mercury manometer is;. 856 mmHg
According to the question;
- The atmospheric pressure is 767 mmHg
- The gauge pressure is 8.9 cmHg = 89mmHg.
The absolute pressure, P(abs);
is given mathematically as;
Absolute pressure = Atmospheric pressure + gauge pressure.
P(abs) = P(atm) + P(gauge)
P(abs) = 767mmHg + 89mmHg.
P(abs) = 856 mmHg.
Read more:
brainly.com/question/17200230
Answer : The volume of gas occupy at
is, 1.25 L
Explanation :
Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.
Mathematically,
![\frac{V_1}{T_1}=\frac{V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BV_1%7D%7BT_1%7D%3D%5Cfrac%7BV_2%7D%7BT_2%7D)
where,
are the initial volume and temperature of the gas.
are the final volume and temperature of the gas.
We are given:
![V_1=1.00L\\T_1=25.0^oC=(25.0+273)K=298K\\V_2=?\\T_2=1.00\times 10^2^oC=((1.00\times 10^2)+273)K=373K](https://tex.z-dn.net/?f=V_1%3D1.00L%5C%5CT_1%3D25.0%5EoC%3D%2825.0%2B273%29K%3D298K%5C%5CV_2%3D%3F%5C%5CT_2%3D1.00%5Ctimes%2010%5E2%5EoC%3D%28%281.00%5Ctimes%2010%5E2%29%2B273%29K%3D373K)
Putting values in above equation, we get:
![\frac{1.00L}{298K}=\frac{V_2}{373K}\\\\V_2=1.25L](https://tex.z-dn.net/?f=%5Cfrac%7B1.00L%7D%7B298K%7D%3D%5Cfrac%7BV_2%7D%7B373K%7D%5C%5C%5C%5CV_2%3D1.25L)
Therefore, the volume of gas occupy at
is, 1.25 L