Answer:
A population or community research line can be carried out, wherever at a certain point in time, regardless of whether it is a cross-sectional study.
In addition, the people who would be the population to be studied or the object of study might or might not know the cause of the study (blind) while the researcher could be experimentally participatory.
Explanation:
They are prevalence studies, in which the presence of a health condition or state is determined in a well-defined population and in a determined time frame: one day, one week, a particular moment in life, even if it does not temporarily coincide in all the subjects (for example, the blood pressure figures at the time of entering the school or at the beginning of the holidays, the prevalence of diabetes in hospitalized patients on a given day, etc.).
They are like "photographs" of a state of affairs at a given moment. The simultaneous determination of what is understood by exposure and event does not allow defining causality.
Answer:
9.8 × 10²⁴ molecules H₂O
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Organic</u>
<u>Stoichiometry</u>
- Analyzing reaction rxn
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[RxN - Unbalanced] CH₄ + O₂ → CO₂ + H₂O
[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O
[Given] 130 g CH₄
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[RxN] 1 mol CH₄ → 2 mol H₂O
[PT] Molar Mass of C: 12.01 g/mol
[PT] Molar Mass of H: 1.01 g/mol
Molar Mass of CH₄: 12.01 + 4(1.01) = 16.05 g/mol
<u>Step 3: Stoichiometry</u>
- [DA] Set up conversion:
- [DA] Divide/Multiply [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
9.75526 × 10²⁴ molecules H₂O ≈ 9.8 × 10²⁴ molecules H₂O
There is no specific name for a glacier that break off as an iceberg. However, the part of the glacier in which this happens is called the "zone of wastage". Chunks break off in a process called "calving".
NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa
NaOH + CH3COOH → CH3COONa + H2O
Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH
Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH
These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L
Molarity of CH3COOH = 0.0106/0.071 = 0.1493M
CH3COONa = 0.0076 / 0.071 = 0.1070M
pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.
pH using Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log ( 0.1070/0.1493)
pH = 4.74 + log 0.717
pH = 4.74 + (-0.14)
pH = 4.60.
It would be NaOH + HCl → <span>NaCl + H2O
</span>
NaOH is sodium hydroxide, which is a strong base. HCl is hydrochloric acid, which is a strong acid.
You have a strong base and a strong acid on the left side, however, at the result side, you end up with NaCl + H2O. Sodium chloride is simply table salt and H2O is just water, thus it has been neutralized.
