Answer:
CH3CHO+H2O → CH3OCH3 - addition
CH,CICH CI + Zn → C2H4 + ZnCl2 - elimination
CH3CH3Br + OH – CH3CH3OH + Br - substitution
2CH2COOH >>(CH3CO)20 + H20 - condensation
Explanation:
An addition reaction is a reaction in which a specie is added across the double bond as we can see in CH3CHO+H2O → CH3OCH3.
In an elimination reaction, a small molecule is lost from a saturated compound to form the corresponding unsaturated compound as in CH,CICH CI + Zn → C2H4 + ZnCl2
In a substitution reaction, a chemical moiety replaces another in a molecule as in; CH3CH3Br + OH – CH3CH3OH + Br .
A condensation reaction is in which two molecules are joined together to form a bigger molecule as in; 2CH2COOH >>(CH3CO)20 + H20.
Answer:
4.62 M
Explanation:
Molarity = moles/volumes (L), so you need to find the moles and the volumes in liters.
Finding the volume is easy because you just have to convert mL to L, so the volume is 0.45 L
Next, find the moles. You can do this by using the molar mass of aluminum to convert the grams to moles. The molar mass of aluminum is 26.98 g/mol.
56 g * (1 mol/26.98 g) = 2.08 mol
Now, divide the moles (2.08) by the volume (.45 L)
Molarity = 4.62 M
Answer:
There are 1.4754246675000002e+24 atoms of Hydrogen within the measurement of 2.45 moles of hydrogen!
Explanation:
Answer:
6626 g
Explanation:
Given that:
Density of water = 1.00 g/ml, volume of water = 42800 ml.
Since density = mass/ volume
mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g
Initial temperature of water = 22°C and final temperature of water = 45°C.
specific heat capacity for water = 4.184 J/g°C
ΔT water = 45 - 22 = 23°C
For iron:
mass = m,
specific heat capacity for iron = 0.444 J/g°C
Initial temperature of iron = 1445°C and final temperature of water = 45°C.
ΔT iron = 45 - 1445 = -1400°C
Quantity of heat (Q) to raised the temperature of a body is given as:
Q = mCΔT
The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.
Q water (gain) + Q iron (loss) = 0
Q water = - Q iron
42800 g × 4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C
m = 4118729.6/621.6
m = 6626 g
