Answer: -
3.3° C
Explanation: -
Mass of water m = 180.5 g
Energy released as heat Q = 2494 J
Specific heat is defined as the heat required to raise the temperature of the unit mass of a given substance by 1 C.
Specific heat of water Cp = 4.184 (J/g)⋅∘C
Using the formula
Q = m x Cp x ΔT
We get temperature change ΔT = Q / (m x Cp)
= 2494 J / ( 180.5 g x 4.184 (J/g)⋅∘C
= 3.3° C
Thus the temprature change, (ΔT), of the wateris 3.3 °C if 180.5 g of water sat in the copper pipe from part A, releasing 2494 J of energy to the pipe
The heat required to vaporize 43.9 g of acetone at its boiling point is calculated as below
the heat of vaporization of acetone at its boiling point is 29.1 kj/mole
find the moles of acetone = mass/molar mass
= 43.9g /58 g/mol =0.757 moles
heat (Q) = moles x heat of vaporization
= 29.1 kj/mole x 0.757 moles = 22.03 kj
<span>Answer: 0.00 meters
Solution:
Step 1: Define displacement
DISPLACEMENT = a vector quantity that describes "linear or angular distance in a given direction between a body or point and a reference position."
Step 2: Understand the question
Assumption 1: Assume that when the ant moves 4.25 meters from its origin to its nest, it is moving in a positive direction (on a graph you would draw a line along the x-axis from its origin to +4.25).
Assumption 2: Assume that when the ant "turns around...back to the source of food", it is moving back in the negative direction (towards the origin).
Step 3: Analyze the question
What is the distance between where the ant originally started and where it ended its journey?
The ant started and ended its journey in the same place.
While it traveled a distance of 8.52 meters (2 * 4.26 = 8.52), it's displacement is actually 0.00 meters (4.26 + (-4.26) = 0.00)
Therefore, the answer is 0.00 meters</span>
Bicarbonate buffer system in blood consists of carbonic acid and bicarbonate ion. H2CO3/HCO3-
When a base enters the body the acid part of the buffer reacts with the base.
Thats the carbonic acid (H2CO3) reacts with the base.
Answer:
There are 0.00523 moles of oxygen in 0.150 grams of calcium sulfate crystal.
Explanation:
Mass of calcium sulfate crystal = m = 0.150 g
Molar mass of calcium sulfate crystal = M = 172 g/mol
Moles of magnesium nitrate = n


1 mole of calcium sulfate crystal has 6 moles of oxygen atoms. Then 0.004446 moles calcium sulfate crystal will have :

There are 0.00523 moles of oxygen in 0.150 grams of calcium sulfate crystal.