Answer:
the stoichiometric coefficient for cobalt is 3
Explanation:
the unbalanced reaction would be
Co(NO₃)₂+ Al → Al(NO₃)₃ + Co
One way to solve is to build a system of linear equations for each element (or group as NO₃) , knowing that the number of atoms of each element is conserved.
For smaller reactions a quick way to solve it can be:
- First the Co as product and as reactant needs to have the same stoichiometric coefficient
- Then the Al as product and as reactant needs to have the same stoichiometric coefficient
- After that we look at the nitrates . There are 2 as reactants and 3 as products . Since the common multiple is 6 then multiply the reactant by 3 and the product by 2.
Finally the balanced equation will be
3 Co(NO₃)₂+ 2 Al → 2 Al(NO₃)₃ + 3 Co
then the stoichiometric coefficient for cobalt is 3
Answer:
NaNO3 (solubility = 89.0 g/100 g H2O)
Explanation:
The solubility of a specie is the amount of solute that will dissolve in one litre of the solvent. Solubility is usually expressed in units of molarity.
Now let us calculate the molarity of the NaNO3 (solubility = 89.0 g/100 g H2O)
Molar mass of NaNO3= 23+14+3(16)= 85gmol-1
Mass of solute=89.0g
Amount of solute= mass of NaNO3/molar mass of NaNO3
Amount of solute= 89.0g/85.0 gmol-1
= 1.0moles of NaNO3
Note that 100g of water=100cm^3 of water.
If 1.0 moles of NaNO3 dissolve in 100cm^3 or water therefore,
x moles of NaNO3 will dissolve in 1000cm^3 of water
x= 1.0 × 1000/ 100
x= 10.0 moles of NaNO3
The answer is A
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Answer:
The answer to this is
The column of water in meters that can be supported by standard atmospheric pressure is 10.336 meters
Explanation:
To solve this we first list out the variables thus
Density of the water = 1.00 g/mL =1000 kg/m³
density of mercury = 13.6 g/mL = 13600 kg/m³
Standard atmospheric pressure = 760 mmHg or 101.325 kilopascals
Therefore from the equation for denstity we have
Density = mass/volume
Pressure = Force/Area and for a column of water, pressure = Density × gravity×height
Therefore where standard atmospheric pressure = 760 mmHg we have for Standard tmospheric pressure= 13600 kg/m³ × 9.81 m/s² × 0.76 m = 101396.16 Pa
This value of pressure should be supported by the column of water as follows
Pressure = 101396.16 Pa = kg/m³×9.81 m/s² ×h
∴ 
= 10.336 meters
The column of water in meters that can be supported by standard atmospheric pressure is 10.336 meters
