A. good conductors - copper, aluminium
b. fair conductors - carbon,human body
c. insulator - paper, wood
948 or 9.48 x 10^2
There are two sets of rules for significant figures
• One set for addition and subtraction
• Another set for multiplication and division
You used the set for multiplication and division.
This problem involves addition and subtraction, and the rule is
The number of places after the decimal point in the answer must be <em>no greater than the number of decimal places in every term</em> in the sum.
Thus, we have
78.9
+890.43
-21.
= 948.33
The "21" term has the fewest digits after the decimal point (none), so the answer must have no digits after the decimal point.
To the correct answer is 948 = 9.48 x 10^2. It has three significant figures.
Explanation:
Acceleration (a) = 600 m/s²
Mass (m) = 300 kg
Force (F) = ?
We know
F = m * a
= 300 * 600
=180000 Newton
The force acting on the object is 18000 Newton.
Explanation:
The given data is as follows.
T =
= (120 + 273.15)K = 393.15 K,
As it is given that it is an equimolar mixture of n-pentane and isopentane.
So,
= 0.5 and
= 0.5
According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.
(393.15 K) = 9.2 bar
(393.15 K) = 10.5 bar
Hence, we will calculate the partial pressure of each component as follows.

= 
= 4.6 bar
and, 
= 
= 5.25 bar
Therefore, the bubble pressure will be as follows.
P =
= 4.6 bar + 5.25 bar
= 9.85 bar
Now, we will calculate the vapor composition as follows.

= 
= 0.467
and, 
= 
= 0.527
Calculate the dew point as follows.
= 0.5,
= 0.5


= 0.101966
P = 9.807
Composition of the liquid phase is
and its formula is as follows.

= 
= 0.5329

= 
= 0.467
Answer:

Explanation:
Formula for the calculation of no. of Mol is as follows:

Molecular mass of Ag = 107.87 g/mol
Amount of Ag = 5.723 g

Molecular mass of S = 32 g/mol
Amount of S = 0.852 g

Molecular mass of O = 16 g/mol
Amount of O = 1.695 g

In order to get integer value, divide mol by smallest no.
Therefore, divide by 0.02657



Therefore, empirical formula of the compound = 