Answer:
70.0°C
Explanation:
We are given;
- Amount of heat generated by propane as 104.6 kJ or 104600 Joules
- Mass of water is 500 g
- Initial temperature as 20.0 ° C
We are required to determine the final temperature of water;
Taking the initial temperature is x°C
We know that the specific heat of water is 4.18 J/g°C
Quantity of heat = Mass × specific heat × change in temperature
In this case;
Change in temp =(x-20)° C
Therefore;
104600 J = 500 g × 4.18 J/g°C × (x-20)
104600 J = 2090x -41800
146400 = 2090 x
x = 70.0479
=70.0 °C
Thus, the final temperature of water is 70.0°C
Answer:
Explanation:
Hello,
In this case, the first step is to compute the volume of the block considering the length, height and width:
Then, we compute the volume in cubic centimetres:
Finally, as the density is given by:
We solve for the mass:
Best regards.
Answer:
i.e. mass of 1 mole of glucose, C6H12O6 = (6 × 12.01 + 12 × 1.01 + 6 × 16.00) g = 180.18 g (using atomic weight data to 2 decimals) 1 mole of carbon atoms weighs 12.01 g and there are 6 moles of C atoms in 1 mole of glucose, so the mass of carbon in 1 mole of glucose = 6 × 12.01 g = 72.06 g.
The answer you are looking for is 4.
Answer:
Explanation:
A student obtains a clean, dry graduated cylinder. She weighs the cylinder and finds the mass to be 32.64 g. She then fills the cylinder with a certain volume of water. She weighs the water-filled cylinder and finds the total mass to be 61.57 g. If the density of the water in the laboratory is 0.9975 g / mL , what is the volume of the water in the graduated cylinder?
