Let's work through each question step-by-step:
### Question 49: How many liters of CO2 gas are contained in 1 kg of CO2 gas at NTP?
Given the possible answers:
a) 22.4 L
b) 100 L
c) 509 L
d) 500 L
To find the correct answer, we should consider the molar volume of an ideal gas at Normal Temperature and Pressure (NTP), which is 22.4 liters per mole for any ideal gas. Since the molecular weight of CO2 is approximately 44 g/mol, we can calculate the number of liters of CO2 in 1 kg (1000 g), which is the correct answer given the choice:
- 1 kg of CO2 gas at NTP contains 509 L.
Thus, the answer is c) 509 L.
### Question 50: Atomic size of element Mg is greater than
Given options are:
a) Si
b) Na
c) Al
d) Both a and c
Atomic size generally decreases across a period (from left to right) and increases down a group in the periodic table.
In the periodic table:
- Magnesium (Mg) is in Group 2.
- Sodium (Na) is in Group 1 and is to the left of Mg.
- Aluminum (Al) is in Group 13 and is to the right of Mg.
- Silicon (Si) is in Group 14 and is also to the right of Mg.
Based on the periodic trends:
- The atomic size of Mg is larger than that of Si (Si is to the right of Mg).
- The atomic size of Mg is larger than that of Al (Al is also to the right of Mg).
Therefore, the correct answer is d) Both a and c.
### Question 1: The difference of temperature of 25°C is equivalent to:
Given options are:
a) 45K
b) 72K
c) 32K
d) 25K
To convert Celsius temperature differences to Kelvin, we use the fact that a change of 1°C is equal to a change of 1K.
Therefore, a temperature difference of 25°C is equivalent to a difference of 25K.
So, the correct answer is d) 25K.
### Question 2: A person holds a bucket of weight 60N. He walks 10 m along the horizontal and then climbs up a vertical distance of 5m. The work done by the man is:
Given options are:
a) 300 J
b) 420 J
c) 720 J
d) None
Work done is calculated by the formula:
[tex]\[ \text{Work Done} = \text{Force} \times \text{Distance} \][/tex]
For horizontal distance, there is no height change, so the work done in walking horizontally is zero because there is no displacement in the direction of the force (downward force, horizontal displacement):
[tex]\[ \text{Work Done horizontally} = 0 \][/tex]
For the vertical distance:
[tex]\[ \text{Work Done vertically} = \text{Force} \times \text{Vertical Distance} \][/tex]
[tex]\[ \text{Work Done vertically} = 60 \, \text{N} \times 5 \, \text{m} \][/tex]
[tex]\[ \text{Work Done vertically} = 300 \, \text{J} \][/tex]
Thus, the total work done by the man is 300 J.
So, the correct answer is a) 300 J.
