The correct order given below shows the changes that occurs in a mice population in response to changes in their environment:
- The population of mice is in an environment with many black rocks
- Mice with black for are more likely to survive and reproduce than mice with brown fur
- After many generations, most of the mice in the population have black fur
- A sandstorm covers most of the population's environment with brown sand
- Mice with black fur are less likely to survive and reproduce than mice with brown fur
- After many generations, most of the mice in the population have brown fur
<h3>What is the correct order for natural selection in the desert environment given?</h3>
Based on the process of natural selection due to envrionmental pressures, the population of the mice in the desert changes as follows before and after the environmental change:
- The population of mice is in an environment with many black rocks
- Mice with black for are more likely to survive and reproduce than mice with brown fur
- After many generations, most of the mice in the population have black fur
- A sandstorm covers most of the population's environment with brown sand
- Mice with black fur are less likely to survive and reproduce than mice with brown fur
- After many generations, most of the mice in the population have brown fur
Therefore, the correct order shows the changes that occurs in a mice population in response to changes in their environment.
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Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
<em />
The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>
The metric unit for mass is B. Kilograms
Hydrogen is the only atom that does not have neutron electrons in the first energy level/shell.
Answer:
4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.
Explanation:
Half life (t1/2) = 8 days
Original mass (No) = 64 g
Elapsed time (t) = 32 days
Mass remaining (Nt) = ?
Using the half life equation we can obtain the mass remaining (Nt)
Nt = No (1/2) ^t/t1/2
Substituting the values, we have;
Nt = 64 * ( 1/2 ) ^32/8
Nt = 64 * (1/2) ^4
Nt = 64 * 0.0625
Nt = 4 g
So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.
