Answer:
the initial temperature of the iron sample is Ti = 90,36 °C
Explanation:
Assuming the calorimeter has no heat loss to the surroundings:
Q w + Q iron = 0
Also when the T stops changing means an equilibrium has been reached and therefore, in that moment, the temperature of the water is the same that the iron ( final temperature of water= final temperature of iron = T )
Assuming Q= m*c*( T- Tir)
mc*cc*(T-Tc)+mir*cir*(T - Tir) = 0
Tir = 20.3 °C + 300 g * 4.186 J/g°C * (20.3 C - 19 °C) / ( 51.9 g * 0.449 J/g°C )
Tir = 90.36 °C
Note :
- The specific heat capacity of water is assumed 1 cal/g°C = 4.186 J/g°C
- We assume no reaction between iron and water
Most plants are called autotrophs they make their own food
The volume of N₂ at STP=56 L
<h3>Further explanation</h3>
Given
2.5 moles of N₂
Required
The volume of the gas
Solution
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, the volume per mole of gas or the molar volume-Vm is 22.4 liters/mol.
So for 2.5 moles gas :
Answer:
1.70 g.cm⁻³
Solution:
Data Given;
Mass = 84.7 g
Volume = 49.6 cm³
Density = ?
Formula Used;
Density = Mass ÷ Volume
Putting values,
Density = 84.7 g ÷ 49.6 cm³
Density = 1.70 g.cm⁻³
