It boils
The solubility increase from 38,7 g KCl/100g water to 40,7 g KCl/100 g water.
The boiling point of chloroform is 61,15 oC.
After 5 hours, the contents of the beaker will remain at 0 °C. The 50 g of ice will not melt since the surrounding water is also at 0 °C, and there is no heat transfer from the environment to raise the temperature. Thus, the beaker will contain both 50 g of ice and 100 g of water, all at the same temperature of 0 °C.
one calorie of heat is able to raise one gram of water one degree Celsius so 400 calories could raise 1g of water 400 degrees, so it would raise the 80g by(400/80) 5 degrees Celsius plus the initial temp of 10 degrees, the 80g of water would have a final temp of 15 degrees Celsius
1 cal/gdegC x 225g x (100 - 50.5)degC heat to the boiling point 540 cal/g x 225g heat to vaporize the water 0.5 cal/gdegC x 225 g x (133-100)degC heat to chang temp of steam = 11137.5 cal + 121500 cal + 3712.5 cal = 14850 cal
The solubility increase from 38,7 g KCl/100g water to 40,7 g KCl/100 g water.
Water is transformed in vapors.
The boiling point of chloroform is 61,15 oC.
After 5 hours, the contents of the beaker will remain at 0 °C. The 50 g of ice will not melt since the surrounding water is also at 0 °C, and there is no heat transfer from the environment to raise the temperature. Thus, the beaker will contain both 50 g of ice and 100 g of water, all at the same temperature of 0 °C.
one calorie of heat is able to raise one gram of water one degree Celsius so 400 calories could raise 1g of water 400 degrees, so it would raise the 80g by(400/80) 5 degrees Celsius plus the initial temp of 10 degrees, the 80g of water would have a final temp of 15 degrees Celsius
The statement that is true is that tar in container A (100 °C) and container B (200 °C) is at a higher temperature than the tar in containers C (50 °C) and D (25 °C). Therefore, the tar in containers A and B is likely to be less viscous and flow more easily compared to the thicker, cooler tar in containers C and D.
1 cal/gdegC x 225g x (100 - 50.5)degC heat to the boiling point 540 cal/g x 225g heat to vaporize the water 0.5 cal/gdegC x 225 g x (133-100)degC heat to chang temp of steam = 11137.5 cal + 121500 cal + 3712.5 cal = 14850 cal
419.1 Joules are required to heat one gram of liquid water from 0.01 degC to 100 deg C. So the answer is 419.1*46 = 19278.6
The mass of undissolved potassium nitrate is cca. 3 g.
Yes, assuming we are talking about pure water and pure ice. Ice is the solid state of water and occurs at 32 degF or 0 degC.
A temperature change of 100 K is equivalent to a change of 100 degrees Celsius. This is because the size of a degree Celsius and a Kelvin is the same, but the zero point is different (0 K is absolute zero while 0°C is the freezing point of water).
Ice is the exception to the rule that most substances become less dense as temperature increases. At 0°C, ice is less dense than liquid water, which is why it floats. In contrast, aluminum, oxygen, and steel all follow the general trend of decreasing density as temperature increases.