The researcher may first weight the beaker with water and then start to heat the water to a constant temperature, for example 30 °C and then start adding salt and stirring. He should add salt slowly until solid salt starts to become visible and the solution starts becoming cloudy. When this happens, he should quickly weigh the beaker. The increase in mass is the mass of salt dissolved at that temperature.
The procedure is then repeated but at an increased temperature until 5-6 temperatures have been tested.
The large piece of jewelry that has a mass of 132.6 g and when is submerged in a graduated cylinder that initially contains 48.6 ml water and the volume increases to 61.2 ml once the piece of jewelry is submerged, has a density of: 10.523 g/ml
To solve this problem the formulas and the procedures that we have to use are:
Where:
- d= density
- m= mass
- v= volume
- v(f) = final volume
- v(i) = initial volume
Information about the problem:
- m = 132.6 g
- v(i) = 48.6 ml
- v(f) = 61.2 ml
- v = ?
- d =?
Applying the volume formula we get:
v = v(f)-v(i)
v = 61.2 ml - 48.6 ml
v = 12.6 ml
Applying the density formula we get:
d = m/v
d = 132.6 g/12.6 ml
d = 10.523 g/ml
<h3>What is density?</h3>
It is a physical quantity that expresses the ratio of the body mass to the volume it occupies.
Learn more about density in: brainly.com/question/1354972
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It's A, t<span>The figure is a molecule and an element.</span>
Answer:
ΔH = -470.4kJ
Explanation:
It is possible to sum 2 or more reactions to obtain the ΔH of the reaction you want to study (Hess's law). Using the reactions:
1. CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(s)ΔH = −414kJ
2. 6C2H2(g) + 3CO2(g) + 4H2O(g) → 5CH2CHCO2H(g)ΔH = 132kJ
6 times the reaction 1.
6CaC2(s) + 12H2O(l) → 6C2H2(g) + 6Ca(OH)2(s)ΔH = −414kJ*6 = -2484kJ
This reaction + 2:
6CaC2(s) + 3CO2(g) + 16H2O(l) → + 6Ca(OH)2(s) + 5CH2CHCO2H(g) ΔH = -2484kJ + 132kJ = -2352kJ
As we want to calculate the net change enthalpy in the formation of just 1 mole of acrylic acid we need to divide this last reaction in 5:
6/5CaC2(s) + 3/5CO2(g) + 16/5H2O(l) → + 6/5Ca(OH)2(s) + CH2CHCO2H(g) ΔH = -2352kJ / 5
<h3>ΔH = -470.4kJ</h3>
