D. Being cold temperatures can result in a cold nose. With prolonged exposure The body will start to lose heat faster than it can generate it, this is the result of hypothermia.
This reaction is decomposition. It is the breakdown of a compound into simpler and smaller elements.
Answer:
2.9 M
Explanation:
Step 1: Given data
Moles of barium chloride (solute): 4.4 moles
Volume of solution: 1.5 liters
Step 2: Calculate the molarity of barium chloride in the solution
The molarity is a way to quantitatively express the concentration of a solute in a solution. The molarity is equal to the moles of solute divided by the volume, in liters, of solution.

Answer:
Water pressure 0.5 atm
Total Pressure= 2.27 atm
Explanation:
To answer this problem, one has to realize that there are two processes that increase the temperature of the sealed vessel.
First, the dry air in the sealed vessel will be heated which will cause its pressure to increase and it can be determined by the equation:
P₁ x T₂ = P₂ x T₁ ∴ P₂ = P₁ x T₂ / T₁
For the second process, we have an amount of n moles of water which will be released when the copper sulfate is heated. In this case, to determine the value of the the water gas we will use the gas law:
PV = nRT ∴ P = nRT/V
n will we calculated from the quantity of sample.
2.50 g CuSo₄ 5H₂O x 1 mol/ 249.69 g = 0.01 mol CuSo₄ 5H₂O
the amount water of hydration is
= 0.01 mol CuSo₄ 5H₂O * 5 mol H₂O / 1 mol CuSo₄ 5H₂O
= 0.05 mo H₂O
pressure of dry air at the final temperature,
P₂ = 1 atm x 500 K/ 300 K = 1.67 atm
Pressure of water :
P (H₂O) 0.05 mol x 0.08206 Latm/kmol x 500 K/ 4 L = 0.5 atm
∴ Total Pressure = 1.67 atm
H2O Pressure = 0.5 atm
Answer:
Specific heat of solid A is greater than specific heat of solid B.
Explanation:
In the calorimeter, as the temperature is increasing, the vibrational kinetic energy will increase and this means that additional amount of energy will be needed to increase the temperature by the same value. Therefore, we can conclude that specific heat increases as temperature increases.
Now, we are told that the final temperature of solid A's calorimeter is higher than that of B.
This means from our definition earlier, Solid A will have a higher specific heat that solid B.