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Answer:
14.91 K.
Explanation:
- To solve this problem, we can use the following relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat transferred to water.
m is the mass of the amount of water (m = 2.0 kg = 2000.0 g).
c is the specific heat capacity of water (c = 4.2 J/g.K).
ΔT is the change in temperature due to the transfer of butane burning.
- To determine Q that to be used in calculation:
Q from 4.000 g of butane is completely burned is - 198.3 kJ = 198300 J.
<em>The negative sign</em><em> symbolizes the the enthalpy change is </em><em>exothermic</em><em>, which means that </em><em>the</em><em> </em><em>energy is released</em><em>.
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- Note that only 63.15% of the energy generated is actually transferred to the water.
∴ Q (the amount of heat transferred to water) = (198300 J)(0.6315) = 125226.45 J.
- Now, we can obtain the change in temperature:
∴ ΔT = Q/m.c. = (125226.45 J) / (2000.0 g)(4.2 J/g.K) = 14.9079 K ≅ 14.91 K.
<em>This means that the temperature is increased by 14.91 K.</em>
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Answer:
pH = 12.58
Explanation Careful
Since NaOH is a strong bond base you have to solve for pOH and then pH
pOH= 1.42
pH= 14-pOH
14-1.42= 12.58
Answer:
a. MgO(s) + H2CO3(aq) → MgCO3(s) + H2O(l) DOUBLE DISPLACEMENT REACTION.
b. 2KNO3(s)→2KNO2 (s) + O2(8) DESCOMPOSITION REACTION.
c. H2(g) + CuO(s) → Cu(s) + H2O(1) SINGLE DISPLACEMENT REACTION.
d. NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(1) COMBUSTION REACTION.
e. H2(8) + Cl2(g) → 2HCl(8) SYNTHESIS REACTION.
f. SO3(g) + H2O(1)→ H2SO4(aq) SYNTHESIS REACTION.