Answer:
C
Explanation:
Usually with these things called Symbiotic Relationships, there is benefit from both parties
The pH of a solution that contains 0.0011 moles of KOH in 393 mL of solution is 2.56. Details about pH can be found below.
<h3>How to calculate pH?</h3>
The pH of a solution can be calculated using the following expression:
pH = - log [H3O+]
However, the hydrogen concentration of the KOH solution can be calculated as follows:
concentration = 0.0011 mol ÷ 0.393 L
concentration = 2.798 × 10-³M
The pH of KOH = - log [2.798 × 10-³M]
pH = 2.56
Therefore, the pH of a solution that contains 0.0011 moles of KOH in 393 mL of solution is 2.56.
Learn more about pH at: brainly.com/question/15289741
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Answer:
a) The mass of lead in the alloy = 0.28 kg
b) There can be dissolved 1.46 kg more of lead.
Explanation:
Step 1: Data given
mass of the magnesium-lead alloy = 5.5 kg
200 °C → 5 wt% Pb
350 °C → 25 wt% Pb
a) What mass of lead is in the alloy?
We have to calculate (for the magnesium-lead alloy) the mass of lead in 5.5 kg of the solid α phase at 200°C just below the solubility limit. The solubility limit for the α phase at 200°C is about 5 wt% Pb.
The mass of lead in the alloy = (0.05)*(5.5 kg) = 0.28 kg
b) If the alloy is heated to 350°C, how much more lead may be dissolved in the α phase without exceeding the solubility limit of this phase?
At 350°C, the solubility limit of the a phase increases to approximately 25 wt% Pb.
C(lead) = ((mass fo lead in alloy) + (mass lead)) / ((magnesium-lead alloy mass) + (mass lead))
0.25 = (0.28 + m(Pb)) / ( 5.5 + m(Pb))
0.25 * ( 5.5 + m(Pb)) = (0.28 + m(Pb))
1.375 + 0.25 m(Pb) = 0.28 + m(Pb)
1.095 = 0.75 m(Pb)
m(Pb) = 1.095 / 0.75
m(Pb) = 1.46
There can be dissolved 1.46 kg more of lead.
5.6L of O2 means we have 0.25 moles of O2.
As, 1 mole has 6.023*10^23 molecules,
0.25 moles of O2 will have 0.25*6.023*10^23 molecules=1.50575*10^23 molecules
and as 1 molecule of O2 has 2 atoms, so, 1.50575*10^23 molecules will have 2*1.50575*10^23 atoms=3.0115*10^23 atoms of O.
Answer:
Through Total internal reflection.
such as formation of multiple images.
Explanation:
As the light ray Strikes the mirror surface, the incident ray is refracted away from the normal and critical angle exceeds the normal.
Hence the light ray partially transmitts or refracts and also reflects.