<h2>The required "option is b) hydrogen bonds must be broken to raise its temperature.</h2>
Explanation:
- Water has high specific heat due to hydrogen bonds present in it.
- The Ionisation of water does not affect the specific heat of the water.
- On decreasing the temperature, there is the formation of bonds hence option (d) is wrong.
- On increasing the temperature, there is the breaking of bonds hence option (b) is correct.
Answer:
0.40 g/cm3
Explanation:
density = mass / volume.
mass = 65.2 grams
volume = 10*1.1*15=165 cm3
so density = 65.2/165=0.40 g/cm3
Answer:
A single molecule of water has been isolated for the first time by trapping it in a fullerene cage. Water molecules are never found alone — they are always hydrogen-bonded to other molecules of water or polar compounds.
While making small volumes of pure water in a lab is possible, it's not practical to “make” large volumes of water by mixing hydrogen and oxygen together. The reaction is expensive, releases lots of energy, and can cause really massive explosions.
While making small volumes of pure water in a lab is possible, it's not practical to “make” large volumes of water by mixing hydrogen and oxygen together. The reaction is expensive, releases lots of energy, and can cause really massive explosions.
A water molecule consists of three atoms; an oxygen atom and two hydrogen atoms, which are bond together like little magnets. The atoms consist of matter that has a nucleus in the centre. The difference between atoms is expressed by atomic numbers.
Explanation:
The balanced chemical equation would be as follows:
<span>K2PtCl4(aq) + 2NH3(aq) --> Pt(NH3)2Cl2(s) + 2KCl(aq)
We are given the amount of </span>K2PtCl4 to be used in the reaction. This will be the starting point for our calculations. We do as follows:
65 g K2PtCl4 ( 1 mol / 415.09 g ) ( 1 mol Pt(NH3)2Cl2 / 1 mol K2PtCl ) ( 300.051 g / 1 mol ) = 46.99 g Pt(NH3)2Cl produced
Answer:
We need 41.2 L of propane
Explanation:
Step 1: Data given
volume of H2O = 165 L
Step 2: The balanced equation
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Step 3: Calculate moles of H2O
1 mol = 22.4 L
165 L = 7.37 moles
Step 4: Calculate moles of propane
For 1 mol C3H8 we need 5 moles O2 to produce 3 moles CO2 and 4 moles H2O
For 7.37 moles H2O we need 7.37/4 = 1.84 moles propane
Step 5: Calculate volume of propane
1 mol = 22.4 L
1.84 moles = 41.2 L
We need 41.2 L of propane
