Answer is: 39,083kJ.
m(coal) = 2,00g.
m(water) = 500g.
ΔT = 43,7°C - 25°C = 18,7°C, <span>difference at temperatures.</span>
c(water) = 4,18 J/g·°C, <span>specific heat of water
</span>Q = m(water)·ΔT·c(water), heat of reaction.
Q = 500g·18,7°C·4,18J/g·°C.
Q = 39083J = 39,083kJ.
Answer:
- The molarity of the student's sodium hydroxide solution is 0.0219 M
Explanation:
<u>1) Chemical reaction.</u>
a) Kind of reaction: neutralization
b) General form: acid + base → salt + water
c) Word equation:
- sodium hydroxide + oxalic acid → sodium oxalate + water
d) Chemical equation:
- NaOH + H₂C₂O₄ → Na₂C₂O₄ + H₂O
b) Balanced chemical equation:
- 2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O
<u>2) Mole ratio</u>
- 2mol Na OH : 1 mol H₂C₂O₄ :1 mol Na₂C₂O₄ : 2 mol H₂O
<u>3) Starting amount of oxalic acid</u>
- mass = 28 mg = 0.028 g
- molar mass = 90.03 g/mol
- Convert mass in grams to number of moles, n:
n = mass in grams / molar mass = 0.028 g / 90.03 g/mol = 0.000311 mol
<u>4) Titration</u>
- Volume of base: 28.4 mL = 0.0248 liter
- Concentration of base: x (unknwon)
- Number of moles of acid: 2.52 mol (calculated above)
- Proportion using the theoretical mole ratio (2mol Na OH : 1 mol H₂C₂O₄)
That means that there are 0.000622 moles of NaOH (solute)
<u>5) Molarity of NaOH solution</u>
- M = n / V (liter) = 0.000622 mol / 0.0284 liter = 0.0219 M
That is the correct number using <em>three signficant figures</em>, such as the starting data are reported.
When winds move air masses, they carry their weather conditions (heat or cold, dry or moist) from the source region to a new region. When the air mass reaches a new region, it might clash with another air mass that has a different temperature and humidity. This can create a severe storm.
Answer:
it is b
Explanation:
A drainage basin is any area of land where precipitation collects and drains off into a common outlet, such as into a river, bay, or other body of water
Answer:
There are 17.64% students received B+ grades.
Explanation:
It is given that,
Total number of students in chemistry class is 17
We need to find the percentage received by B+.
Number of students having B+ grades are 3 (from graph)
Required percentage =
So, there are 17.64% students received B+ grades.