Answer:
57.6g
Explanation:
So, if in one mole of water, 16 g of oxygen atom is present. Then, in 3.6 moles of water, the mass of oxygen present will be 3.6×16=57.6g. Therefore, the amount of oxygen present in 3.6 g water is option (B)- 57.6 g.
Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
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Answer:
Al(NO3)3(s)--------> Al^3+(aq) + 3NO3^-(aq)
Explanation:
The equation shown above describes the dissolution of Al(NO3)3 in water using the lowest coefficients.
This occurs when solid Al(NO3)3 is added to water. It dissolves to give rise to ions as shown. This is a property of all ionic substances.
I think it’s classified as a sugar. The ending -ose usually means a sugar.
The nonmetal elements have a negative charge.