V=2.90 L
c=1.8 mol/L
n(NaOH)=vc
n(NaOH)=2.90L*1.8mol/L=5.22 mol
Answer:
90 °C
Explanation:
First, we must know the specific heat capacity of water, which is defined as the energy required to heat 1 gram of water by one degree Celsius. The specific heat capacity of water is 1 cal·g⁻¹°C⁻¹.
The equation we will use is Q = mcΔt, where Q is the heat energy, m is the mass, c is the specific heat capacity, and Δt is the temperature change. We will rearrange the equation to solve for Δt and substitute the values:
Δt = Q / (mc) = (90 kcal)(1000 cal/kcal) / (1 kg)(1000 g/kg)(1 cal·g⁻¹°C⁻¹) = 90 °C
Reaction: 2K₍s₎ + 2H₂O₍l₎ → 2KOH₍aq₎ + H₂₍g₎.
K - potassium.
H₂O - water.
KOH - potassium-hydroxide.
H₂ - hydrogen.
s - solid phase.
l - liquid.
aq - disolves in water.
g - gas.
Reaction is exothermal (release of energy) and potassium burns a purple flame. H<span>ydrogen released during the reaction reacts with </span>oxygen<span> and ignites.</span><span>
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