Consider this balanced chemical equation:
2 H2 + O2 → 2 H2O
We interpret this as “two molecules of hydrogen react with one molecule of oxygen to make two molecules of water.” The chemical equation is balanced as long as the coefficients are in the ratio 2:1:2. For instance, this chemical equation is also balanced:
100 H2 + 50 O2 → 100 H2O
This equation is not conventional—because convention says that we use the lowest ratio of coefficients—but it is balanced. So is this chemical equation:
5,000 H2 + 2,500 O2 → 5,000 H2O
Again, this is not conventional, but it is still balanced. Suppose we use a much larger number:
12.044 × 1023 H2 + 6.022 × 1023 O2 → 12.044 × 1023 H2O
These coefficients are also in the ratio of 2:1:2. But these numbers are related to the number of things in a mole: the first and last numbers are two times Avogadro’s number, while the second number is Avogadro’s number. That means that the first and last numbers represent 2 mol, while the middle number is just 1 mol. Well, why not just use the number of moles in balancing the chemical equation?
2 H2 + O2 → 2 H2O
<span>the chemical equation will be Ni(OH)2(s)------>heat---> NiO(s) + H2O(g)
</span><span>we know that the heat supplied to decompose the compound.In the result the product H2O is assumed to be in the vapor state so that is gas.
hope it helps
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98 elements are naturally forming elements.
Answer: 98
Answer:
Qm = -55.8Kj/mole
Explanation:
NaOH(aq) + HNO₃(aq) => NaNO₃(aq) + H₂O(l)
Qm = (mc∆T)water /moles acid
Given => 100ml(0.300M) NaOH(aq) + 100ml(0.300M)HNO₃(aq)
=> 0.03mole NaOH(aq) + 0.03mole HNO₃(aq)
=> 0.03mole NaNO₃(aq) + 0.03mole H₂O(l)
ΔH⁰rxn = [(200ml)(1.00cal/g∙°C)(37 – 35)°C]water / 0.03mole HNO₃
= 13,333 cal/mole x 4.184J/cal = 55,787J/mol = 55.8Kj/mole (exothermic)*
Heat of reactions comes from formation of H-Oxy bonds on formation of water of reaction and heats the 200ml of solvent water from 35⁰C to 37⁰C.
