First of all we should know that, 1 Joule = 0.000239 kilocalories.
So, 6.6×
J = 6.6×
× 0.000239 kilocalories
6.6×
J = 15774000 kilocalories
= 1.58 ×
kilocalories
One joule is described as the quantity of electricity exerted when a pressure of 1 newton is implemented over a displacement of 1 meter. Within the SI machine, the unit of labor or electricity is the Joule.
The kilocalorie, or meals calorie, is the quantity of warmth required to elevate one kilogram of water 1 °C. warmness capability is the amount of heat required to raise one gram of material 1 °C beneath steady pressure. A kilocalorie is the amount of warmth required to raise the temperature of 1 kilogram of water one diploma Celsius.
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Answer: The density of silver metal will be ![10.50g/ml[\tex]Explanation:Density is defined as the mass contained per unit volume.[tex]Density=\frac{mass}{Volume}](https://tex.z-dn.net/?f=10.50g%2Fml%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3EExplanation%3A%3C%2Fp%3E%3Cp%3EDensity%20is%20defined%20as%20the%20mass%20contained%20per%20unit%20volume.%3C%2Fp%3E%3Cp%3E%5Btex%5DDensity%3D%5Cfrac%7Bmass%7D%7BVolume%7D)
Given : Mass of silver = 194.3 grams
Volume of silver= volume of water displaced= ![260.5-242.0=18.5mltex]Putting in the values we get:[tex]Density=\frac{194.3g}{18.5ml}=10.50g/ml/tex]Thus density of silver metal will be [tex]10.50g/ml](https://tex.z-dn.net/?f=260.5-242.0%3D18.5mltex%5D%3C%2Fp%3E%3Cp%3EPutting%20in%20the%20values%20we%20get%3A%3C%2Fp%3E%3Cp%3E%5Btex%5DDensity%3D%5Cfrac%7B194.3g%7D%7B18.5ml%7D%3D10.50g%2Fml%2Ftex%5D%3C%2Fp%3E%3Cp%3EThus%20density%20of%20silver%20metal%20will%20be%20%5Btex%5D10.50g%2Fml)
50.0ml is the answer nacho
Hey there!:
mass = 41.2 g
Volume = 8.2 cm³
Therefore:
D = m / V
D = 41.2 / 8.2
D = 5.02 g/cm³
Energy absorbed from
changing the phase of a substance from the liquid phase to gas phase can be
calculated by using the specific latent heat of vaporization. Calculation are
as follows:
<span> Energy = 1.35 mol CCl4 ( 153.81 g / 1 mol ) x 197.8 J/g </span>
<span>Energy = 41071.88 J</span>