Answer:
1.0 ° C
Explanation:
The molar mass for Sodium Nitrate NaNO₃ = (23+14+(16×3)) = 85
Number of moles of NaNO₃ = mass of NaNO₃ /molar mass of NaNO₃
⇒ 17/85 = 1.38 moles
Since 1 mole of NaNO₃ dissolved in 1 cubic decimeter of water, 40 kJ of heat energy is absorbed.
when 1.38 mole of NaNO₃ dissolved in 1 cubic decimeter of water, x kJ of heat energy is absorbed..
Then; x kJ of 1.38 mole of NaNo₃ = 1.38 × 40 kJ =55.2 kJ of heat absorbed.
Using the relation : Q = mcΔT to determine the temperature drop ; we get:
55.2 = 17 × 4 (ΔT)
55.2 = 68 ΔT
ΔT= 0.8 ° C
ΔT ≅ 1.0 ° C
Therefore, the drop in temperature when 17.0g of sodium nitrate is dissolved in 1 cubic decimeter of water is 1.0 ° C
I need the options to choose from
Explanation:
As it is known that in solids, molecules are held together because of strong intermolecular forces of attraction. As a result, they are held together and have definite shape and volume.
Whereas in liquids, molecules are not held so strongly as they are in solids. Hence, they move from their initial position and they do not have definite shape but they have definite volume.
Liquids obtain the shape of container in which they are kept.
In gases, molecules are held together by weak intermolecular forces. As a result, they move far apart from each other and occupy the space of a container or vessel in which they are placed.
The physical state (at room temperature) of the following are determined as follows:
(a) Helium in a toy balloon : Helium at room temperature exists as a gas. So, when helium is present in a toy balloon then it acquires the volume of toy balloon.
(b) Mercury in a thermometer : Mercury at room temperature exists as a liquid. When it is placed in a thermometer then volume of mercury does not get affected.
(c) Soup in a bowl : Since, soup is a liquid. Hence, its volume will not change according to the volume of container.
Answer:
120 g of NaCl in 300 g H20 at 90 C
Explanation:
At x = 90 go vertical to the line for NaCl...then go left to the y-axis to find the solubility in 100 g H20 = 40
we want 300 g H20 so multiply this by 3 to get 120 gm of NaCl in 300 g
Staining specimen with heavy metal salts (e.g. tungsten, molybdenum) allows you to see the specimen better with higher contrast when electron beam deflects off of your sample.
