It’s DEFINITELY 2 like DEFINITELY
Answer:
The heat capacity for the sample is 0.913 J/°C
Explanation:
This is the formula for heat capacity that help us to solve this:
Q / (Final T° - Initial T°) = c . m
where m is mass and c, the specific heat of the substance
27.4 J / (80°C - 50°C) = c . 6.2 g
[27.4 J / (80°C - 50°C)] / 6.2 g = c
27.4 J / 30°C . 1/6.2g = c
0.147 J/g°C = c
Therefore, the heat capacity is 0.913 J/°C
Answer:
15.2 g H2
Explanation:
2H2O -> 2H2 + O2
9.06 x 10^24 molecules x (1 mol/6.022 x 10^23 molecules) x (2 mol H2/2 mol H2O) x (1.008 g/1 mol) = 15.2 g H2
Answer:
106.25 mL
Explanation:
For this, we can use
C1×V1=C2×V2
C1 = 0.45
V1 = 85
C2= 0.20
V2= ?
0.45 × 85 = 0.20 × V2
V2= (0.45 × 85)/0.20
V2=191.25mL
To find the amount of water added, subtract V1 from V2
191.25 - 85 =106.25mL
Explanation:
Ion-dipole interactions are defined as the interactions that occur when an ion interacts with the dipole of a molecule (polar molecule).
For example, 
, 
, 
, NaCl etc are all the molecules that will show ion-dipole interactions.
Dipole-dipole interactions are defined as the interactions that occur when partial positive charge on an atom is attracted by partial negative charge on another atom.
When a polar molecules produces a dipole on a non-polar molecule through distribution of electrons then it is known as dipole-induced forces.
For example, 
, 
etc are the molecules which show dipole-dipole interactions.
