Answer:
The rusting of iron is spontaneous at low temperatures.
Explanation:
The given chemical reaction is:
4Fe(s) + 3O2(g) ----> 2Fe2O3(s) [rust]
The rusting of iron is a chemical reaction in which iron reacts with oxygen in presence of moisture and forms iron oxide.
This reaction takes place in a faster rate when there is low temperatures in the atmosphere.
When temperature is low, the moisture in the atmosphere is more and hence, rate of rusting is more.
<span> reason is that there is no land to slow down the wind. Also, wind is caused by differences in air pressure</span>
Answer:
Because time is independent of distance, and distance is dependent of time.
Explanation:
Usually, on any graph, the independent variable is plotted on the x-axis and the dependent variable is plotted on the y-axis. Because of this, time, which is independent (time happens regardless of any other factor), is on the x-axis while distance, which is the dependent variable (can only take place in time), is on the y-axis.
9ml will be given for the case of dosage calculation order: 3 mg available: 2 mg per 6 ml
Conversion factors are necessary for dosage calculation, such as when translating from pounds to kilograms or liters to milliliters. This approach, which is straightforward in design, enables physicians to deal with different units of measurement and convert factors to arrive at the solution.
dosage calculation techniques serve as a second or third check on the accuracy of the previous computation techniques. Dimensional Analysis, Ratio Proportion, and Formula or Desired Over Have Method are the three main approaches for dosage calculation. dosage calculations are frequently prescribed and labeled based on their weight or, for solutions, their strength, which is the amount of weight dissolved or suspended in a given volume.
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Answer:
1 x 10⁻¹¹ M
Explanation:
<u>(Step 1)</u>
Determine the pH.
pH = -log[H⁺]
pH = -log[1 x 10⁻³ M]
pH = 3
<u>(Step 2)</u>
Determine the pOH.
pH + pOH = 14
3 + pOH = 14
pOH = 11
<u>(Step 3)</u>
Determine the hydroxide (OH⁻) concentration.
[OH⁻] = 10^-pOH
[OH⁻] = 10⁻¹¹
[OH⁻] = 1 x 10⁻¹¹ M
