Explanation:
To solve this problem, we need to understand the concept of precision and accuracy.
Through this understanding we apply to solve this problem.
- Precision is the ability to reproduce the same set of values in an experiment.
- Accuracy is the nearness or closeness of the measured value to the true value.
For example; if the true value is 28.0mm and the following readings were made:
A = 24.0mm, 24.0mm, 24.3mm, 23.9mm This is a precise measurement but not accurate.
B = 27.9mm, 28.0mm, 28.1mm: This is an accurate measurement because it is close to the true value.
Answer:
percentage by mass of Ba = 0.4583/0.8360 × 100 = 45.83
/0.8360 = 54.82%
Explanation:
The mass of Barium in the original compound is the same as the masss of barium in the precipitate BaSO4.
Ba²+
molar mass of BaSO4 = 137 + 32 + 16(4) = 137 + 32 + 64 = 233 g/ mol
since
137 grams of barium is in 233 g of BaSO4
? grams of barium is in 0.7794 g of BaSO4
cross multiply
grams of barium = 0.7794 × 137/233
grams of barium = 106.7778/233
grams of barium = 0.45827381974
grams of barium = 0.4583 grams
percentage by mass of Barium in the original compound =
mass of barium/ mass of the compound × 100
mass of barium = 0.4583 g
mass of compound = 0.8360 g
percentage by mass of Ba = 0.4583/0.8360 × 100 = 45.83
/0.8360 = 54.82%
Answer:
No component is perfect. All have tolerances that can vary. If you construct a simple circuit where a 10 volt power supply feeds a 10 ohm resistor, you would expect to measure a current of one ampere. BUT - the wiring has some resistance too. This adds perhaps 0.1 ohms to the circuit. The resistor has a +-5% tolerance. If it is 5% high, it may measure 10.5 ohms. That's a total circuit resistance of 10.6 ohms. The power supply may have a tolerance of +-1%. Suppose it's 1% low. That's an output of 9.9 volts in real life. So you have 9.9 volts dropped across 10.6 ohms. you will measure closer to 0.934 amps instead of 1.000 amps. To make matters worse, most electronic components have a temperature coefficient, that is, their values change with different temperatures. You may get a completely different reading tomorrow if the temperature is different! Finally, with current measurements in particular, you are inserting the ammeter in series with the circuit under test. Ammeters have some inherent resistance too, so by putting the ammeter in the circuit, you are changing the very current you are trying to measure (a little)! Oh yeah, the ammeter has a tolerance too. Its reading may be off a little even if everything else is perfect. Sometimes you have to wonder how we get a decent reading at all. Fortunately the errors are usually fairly small, and not all tolerances are off in the same direction or off the maximum amount. They tend to cancel each other out somewhat. BUT - in rare circumstances everything CAN happen like I said, and the error can be huge.
Explanation:
Answer:
Volume occupied by Neon gas is 52.67 L
Explanation:
Using Ideal Gas Equation:
PV = nRT
where
P = pressure exerted by the gas = 57 atm
V = volume occupied = ?
n = number of moles = 115 moles
R = Ideal gas constant = 0.0821 L.atm/K.mol
(R value should be taken according to the units of Temperature,pressure, volume and mole)
T = Temperature = 45 + 273 = 318 K
(For temperature conversion from C to K add 273 to temperature:T + 273)
PV = nRT , So
Put values of T,P,n,R
V = 52.67 L
Volume occupied by 115 moles of Neon gas at 57 atm Pressure and 45 C temperature is 52.67 L
