Since you didn't have any extra information about the question I'll be presenting an example from my own textbooks that I've used.
An example of a direct observation is listening to a cricket chirp at night, and counting the number of chirps per minute.
Direct Observation is where the evaulator watches the subject in their usual habitat without disrupting or altering it.
Answer:
378mL
Explanation:
The following data were obtained from the question:
Pressure (P) = 99.19 kPa
Temperature (T) = 28°C
Number of mole (n) = 0.015 mole
Volume (V) =...?
Next, we shall convert the pressure and temperature to appropriate units. This is illustrated below:
For Pressure:
101.325 KPa = 1 atm
Therefore, 99.19 kPa = 99.19/101.325 = 0.98 atm
For Temperature:
T(K) = T(°C) + 273
T(°C) = 28°C
T(K) = 28°C + 273 = 301K.
Next we shall determine the volume of N2. The volume of N2 can be obtained by using the ideal gas equation as shown below:
PV = nRT
Pressure (P) = 0.98 atm
Temperature (T) = 301K
Number of mole (n) = 0.015 mole
Gas constant (R) = 0.0821atm.L/Kmol.
Volume (V) =...?
0.98 x V = 0.015 x 0.0821 x 301
Divide both side by 0.98
V = (0.015 x 0.0821 x 301) /0.98
V = 0.378 L
Finally, we shall convert 0.378 L to millilitres (mL). This is illustrated below:
1L = 1000mL
Therefore, 0.378L = 0.378 x 1000 = 378mL
Therefore, the volume of N2 collected is 378mL
Answer:
27.60 g urea
Explanation:
The <em>freezing-point depression</em> is expressed by the formula:
In this case,
- ΔT = 5.6 - (-0.9) = 6.5 °C
m is the molality of the urea solution in X (mol urea/kg of X)
First we<u> calculate the molality</u>:
- 6.5 °C = 7.78 °C kg·mol⁻¹ * m
Now we<u> calculate the moles of ure</u>a that were dissolved:
550 g X ⇒ 550 / 1000 = 0.550 kg X
- 0.84 m = mol Urea / 0.550 kg X
Finally we <u>calculate the mass of urea</u>, using its molecular weight:
- 0.46 mol * 60.06 g/mol = 27.60 g urea
Answer:
See attached picture.
Explanation:
Hello!
In this case, since C2H3Cl is an organic compound we need a central C-C parent chain to which the three hydrogen atoms and one chlorine atom provides the electrons to get all the octets except for H as given on the statement.
In such a way, on the attached picture you can find the required Lewis dot structure without formal charges and with all the unshared electron pairs, considering there is a double bond binding the central carbon atoms in order to compete their octets.
Best regards!
I believe your answer is D