Answer:
1. 31.25 mL
2. 1.98 g/L
3. 0.45 g/mL
Explanation:
For each of the problems, you need to perform unit conversions. You need to use the information given to you to convert to a specific unit.
1. You need volume (mL). You have density (g/mL) and mass (g). Divide mass by density. You will cancel out mL and be left with g.
(50.0 g)/(1.60 g/mL) = 31.25 mL
2. You are given grams and liters. You need to find density with units g/L. This means that you have to divide grams by liters.
(0.891 g)/(0.450 L) = 1.98 g/L
3. You have to find density again but this time with units g/mL. Divide the given mass by the volume.
(10.0 g)/(22.0 mL) = 0.45 g/mL
Answer:
F = 50 N
Explanation:
Given data:
Mass of car = 250 Kg
Acceleration of car = 0.20 m/s²
Force required = ?
Solution:
Formula:
F = m×a
F = applied force
m = mass
a = acceleration
Now we will put the values in formula.
F = 250 Kg × 0.20 m/s²
F = 50 Kg.m/s²
Kg.m/s² = N
F = 50 N
Answer:
Explanation:
The triple point of carbon dioxide is 5.11 atmosphere at -56.6 degree celsius . At pressure greater than 5.11 , solid carbon dioxide liquefies , as it is warmed. At pressure lesser than 5.11 atmosphere , it will go into gaseous state without liquefying . Excessive pressure helps liquification process.
So maximum pressure required is 5.11 atmosphere. Beyond this pressure , solid CO2 will liquify.
