The answer to this question would be: too low
Molar mass would be determined by the number of mol and the mass of the object. Mass wouldn't be influenced by the temperature, but number of mol is. Using ideal gas formula of PV=nRT you can conlude that the amount of mol(n) is inversely related to the temperature (T).
If the temperature is higher than it supposed to be, then the amount of mol would be lower than it supposed to be.
Acetone will react with ammonia. Acetone will act as an Lewis Acid and ammonia will act as a Lewis Base in this situation. The nitrogen in NH3 is very reactive because of its lone pair of electrons(The nitrogen is delta -ve), so it will attack the carbon cation center of the acetone (which is delta +ve).
Ah , a cup of hot chocolate is alot of chocolate. Im gonna drool ; )
Well , heat flows from an area of high temperature to an area of low temperature. Here , hot chocolate has the high temp , and the surrounding has a room temp. So , the heat from the hot chocolate will dissipate into the surroundings and create a thermal equilibrium. So youre right.
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Answer:
Option A.
2Na + 2H2O —> 2NaOH + H2
Explanation:
To know which option is correct, we shall do a head count of the number of atoms present on both side to see which of them is balanced. This is illustrated below below:
For Option A:
2Na + 2H2O —> 2NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 2 Na
4 H >>>>>>>>>>>> 4 H
2 O >>>>>>>>>>>> 2 O
Thus, the above equation is balanced.
For Option B:
2Na + 2H2O —> NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 1 Na
4 H >>>>>>>>>>>> 3 H
2 O >>>>>>>>>>>> 1 O
Thus, the above equation is not balanced.
For Option C:
2Na + H2O —> 2NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 2 Na
2 H >>>>>>>>>>>> 4 H
1 O >>>>>>>>>>>> 2 O
Thus, the above equation is not balanced.
For Option D:
Na + 2H2O —> NaOH + 2H2
Reactant >>>>>>> Product
1 Na >>>>>>>>>>> 1 Na
4 H >>>>>>>>>>>> 5 H
2 O >>>>>>>>>>>> 1 O
Thus, the above equation is not balanced.
From the illustrations made above, only option A is balanced.
