Answer:
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Explanation:
<u>Step 1: </u>Data given
mass of water = 300 grams
initial temperature = 10°C
final temperature = 50°C
Temperature rise = 50 °C - 10 °C = 40 °C
Specific heat capacity of water = 4.184 J/g °C
<u>Step 2:</u> Calculate the heat
Q = m*c*ΔT
Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)
Q = 50208 Joule = 50.2 kJ
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Answer:
The attractive forces must be overcome are :
Explanation:
For the compound to dissolve the attractive forces existing between atoms of the compound must be reduced
<u>CsI is ionic compound </u><em>and its molecules are held together by ionic(electrostatic) force . These force must be weakened for its dissolution</em>
Forces in HF <em>:</em>
<em>1 .Hydrogen Bonding : In HF strong intermolecular Hydrogen Bonding exist between the electronegative F and Hydrogen</em>
2. Dipole - dipole : <em>HF is polar . So it is a permanent dipole and has dipole diople interaction</em>
Answer:
C: The temperature of the substance increases as it sits in the beaker of water
Explanation:
This question was taken from a video where an attempt was made to investigate the changes in temperature when a substance undergoes change from it's solid phase to its liquid phase.
To do this, as seen in the video online, it shows a solid substance in a test tube being placed in a beaker of water.
From observation, the water in the beaker has a warmer temperature than the solid substance present in the test tube and this in turn makes the test tube gradually increase in temperature.
Thus, the solid substance will as well increase increase in temperature when it is placed in the beaker of water.
Answer:
1.8 × 10² s
Explanation:
Let's consider the reduction that occurs upon the electroplating of copper.
Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)
We will establish the following relationships:
- 1 g = 1,000 mg
- The molar mass of Cu is 63.55 g/mol
- When 1 mole of Cu is deposited, 2 moles of electrons circulate.
- The charge of 1 mole of electrons is 96,486 C (Faraday's constant).
- 1 A = 1 C/s
The time that it would take for 336 mg of copper to be plated at a current of 5.6 A is:
Answer:
0.24 M
Explanation:
Molarity = Moles solute / Liters solution
Step 1: Identify variables
400 mL = Liters solution
0.60 moles = Moles solute
Step 2: Identify conversions
1 L = 1000 mL
Step 3: Convert mL to L
400mL(1 L/1000mL) = 0.4 L
Step 4: Find molarity
M = (0.4 L)(0.60 mol) = 0.24 M
