1)
Answer:
0.00421
Explanation:
Significant figures:
The given measurement have four significant figures 1234.
All non-zero digits are consider significant figures like 1, 2, 3, 4, 5, 6, 7, 8, 9.
Leading zeros are not consider as a significant figures. e.g. 0.03 in this number only one significant figure present which is 3.
Zero between the non zero digits are consider significant like 104 consist of three significant figures.
The zeros at the right side e.g 2400 are also significant. There are four significant figures are present.
In given value the 0.00 are not significant.
when we round to three significant figures we must include 4207. The number after the third significant figure which 7 is greater than 5 so we will increase the third significant figure by one for example 0 become 1.
If the number which is next to rounded figure is 5 or grater than five the rounded figure will increase by 1.
2) In which of following pairs of numbers does each member of the pair contain the same number of significant figures?
a) 0.0806 and 960,000
b) 71.0 and 81.00
c) 20.0 and 462
d) 0.03070 and 0.0670
Answer:
c) 20.0 and 462
Explanation:
The option c is correct because both value have same number of significant figure which are 3.
All other are incorrect because,
a) 0.0806 and 960,000
0.0806 have three significant figures and 960,000 have six significant figures .
b) 71.0 and 81.00
71.0 have three significant figures and 81.00 have four significant figures .
d) 0.03070 and 0.0670
0.03070 have four significant figures and 0.0670 have three significant figures .
3) The mathematical meaning associated with the metric system prefixes milli , micro and nano is,
a) 10⁻², 10⁻⁴ and 10⁻⁶
b) 10⁻², 10⁻³ and 10⁻⁶
c) 10⁻³, 10⁻⁹ and 10⁻¹²
d) 10⁻³, 10⁻⁶ and 10⁻⁹
Answer:
d) 10⁻³, 10⁻⁶ and 10⁻⁹.
Explanation:
The correct option is d.
The mili is represented by "m". In metric system it is unit prefix represent thousand as 10⁻³.
This unit is adopted in 1795.
The micro is Greek word. It is represented by symbol μ .
It is equal to 10⁻⁶.
It is presented in 1960.
The nano is equal to 10⁻⁹. It is represented by symbol "n" .
one nanometer is equal to 10⁻⁹ meter.
All other options are incorrect.
4) In which of the following measured numbers are all of zero significant?
a) 0.00480
b)0.300120
c) 6.080110
d) 99400
Answer:
option c is correct.
Explanation:
a) 0.00480
There are three significant figure in given value 480. The three zeros are not significant.
b)0.300120
There are six significant figure in given value 300120. The zero before decimal point is not significant.
c) 6.080110
There are seven significant figure in given value 6,0,8,0,1,1,0. The all zeros are significant.
d) 99400
There are three significant figure in given value 9,9,4.
5)When a substance undergoes a physical change it is always true that
a) it changes color
b) it changes state
c heat is absorbed
d) its chemical composition changes
Answer:
b) it changes state
Explanation:
Physical Change
The changes that occur only due to change in shape or physical properties but their chemical or internal composition remain unchanged.
These changes were reversible.
They have same chemical property.
These changes can be observed with naked eye.
Example :
Water converting to Ice
Water converting to gas
In this water molecule remain the same only they rearrange themselves that change its state of mater not composition
6) Object A weighs 40 grams and has a volume of 5 mL. Object B weighs 12 grams and has a volume.
Therefore,
a) A and B have equal densities.
b) B is twice as dense as A.
c) B is four times as dense as A.
d) A is more dense than B.
Answer:
Explanation:
Density:
Density is equal to the mass of substance divided by its volume.
Units:
SI unit of density is Kg/m3.
Other units are given below,
g/cm3, g/mL , kg/L
Formula:
D=m/v
D= density
m=mass
V=volume
Symbol:
The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.
Given data:
For object A:
Mass = 40 g
Volume = 5 mL
Density = ?
Solution:
d = m/ v
d = 40 g/ 5 mL
d = 8 g/mL
The volume of object b is missing we can not determine density by just mass value.
.
.