Answer:
- <em>The relationship between the scale divisions marked on the graduated cylinders and the estimated uncertainty in the volume measurements is</em> that <u>the larger the divisions the larger the uncertainty, and the smaller the divisions the smaller the uncertaintity.</u>
Explanation:
In general, <em>graduated cylinders</em> are marked with divisions (lines) at equal spaces or intervals. There are divisions that show the measure with a number and a series of intermediate not numbered divisions in between two numbered lines.
The <em>uncertainty </em>of the instrument is determined as half the difference between two consecutive marks.
So, the larger the interval of two consecutive marks the larger the uncertainty.
The attached image, taken from brainly.com/question/13102373, shows you some marks in a graduated cylinder.
The number at the bottom of each element on the periodic table represents the weight and or atomic weight.
Water and copper hope this helps:)
Answer:
0.342 m
Explanation:
From the question given above, the following data were obtained:
Mass of NaBr = 14.57 g
Mass of water = 415 g
Molar mass of NaBr = 102.89 g/mol
Molality of NaBr =?
Next, we shall determine the number of mole in 14.57 g of NaBr. This can be obtained as follow:
Mass of NaBr = 14.57 g
Molar mass of NaBr = 102.89 g/mol
Mole of NaBr =?
Mole = mass / molar mass
Mole of NaBr = 14.57 / 102.89
Mole of NaBr = 0.142 mole
Next, we shall convert 415 g of water to kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
415 g = 415 g × 1 Kg / 1000 g
415 g = 0.415 Kg
Thus, 415 g is equivalent to 0.415 Kg.
Finally, we shall determine Molality of the solution as follow:
Mole of NaBr = 0.142 mole
Mass of water = 0.415 Kg
Molality of NaBr =?
Molality = mole / mass of water in Kg
Molality of NaBr = 0.142 / 0.415
Molality of NaBr = 0.342 m
Therefore, the molality of NaBr solution is 0.342 m.
Answer: 996 mmHg
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number 
of particles.
According to the ideal gas equation:
P = Pressure of the gas = ?
V= Volume of the gas = 25.5 L
T= Temperature of the gas = 13°C = (273+13) K = 286K
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas= 1.42
(760mmHg=1atm)
Thus pressure of this gas sample is 996 mm Hg.
