The experiment involving the determination of the number of ice cubes required to keep the temperature of the glass under 15 degrees Celcius, the following things have to be kept in mid:
- The<u> temperature</u> of the surroundings
- The initial temperature of the <u>glass</u>
- The <u>number of ice cubes </u>added to the water in the glass
In order to keep into consideration the changing environmental temperatures (which is a variable in the experiment), the experiment had to be conducted daily to get <u><em>accurate results </em></u>keeping into consideration all the factors.
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Answer:
Percent Yield = 94.237%
Explanation:
CO = Carbon Dioxide = Molar Mass 28g/mol
C = Carbon = 12g/mol
O = Oxygen = 16g/mol
Theoretical yield = 93.7 grams
Actual yield = 88.3 grams
Percent yield =
(actual yield
/theoretical yield
)x100
Percent Yield = (88.3/93.7)x100
Percent Yield = 94.237%
Answer:
19 g
Explanation:
Data Given:
Sodium Chloride (table salt) = 50 g
Amount of sodium (Na) = ?
Solution:
Molecular weight calculation:
NaCl = 23 + 35.5
NaCl = 58.5 g/mol
Mass contributed by Sodium = 23 g
calculate the mole percent composition of sodium (Na) in sodium Chloride.
Since the percentage of compound is 100
So,
Percent of sodium (Na) = 23 / 58.5 x 100
Percent of sodium (Na) = 39.3 %
It means that for ever gram of sodium chloride there is 0.393 g of Na is present.
So,
for the 50 grams of table salt (NaCl) the mass of Na will be
mass of sodium (Na) = 0.393 x 50 g
mass of sodium (Na) = 19 g
Answer: Temperature = T, unknown
Saturated Solution, NH4Cl concentration = 60g/100g H2O = 0.6g NH4Cl/g H2O
Assume density of H2O = 1 g/ml
m = 0.6g NH4Cl/g H2O / 1 g/ml
m = 0.6g NH4Cl/ml
See the table of saturated solutions and identify the temperature at which the concentration of NH4Cl is 60g/100g H2O.
Explanation: The line on the graph on reference table G indicates a saturated solution of NH4CL as a concentration of 60. g NH4 Cl/100. g H2O
