Answer:
The answer to your question is Cm = 2.093 J/kg°C
Explanation:
Data
mass metal = mass of water
Δ water = 5°
Δ metal = 10°
Cm = ?
Cw = 4.186 J/Kg°C
Formula
for metal for water
mCmΔTm = mCwΔTw
Cancel mass of metal
Cm ΔTm = CwΔTw
solve for Cm
Cm = (CwΔTw) / ΔTm
Substitution
Cm = (4.186 x 5) / 10
Simplification
Cm = 20.93 / 10
Result
Cm = 2.093 J/kg°C
Answer and explanation:
A) An ideal fuel must:
- easy to transport and storage.
- have a high calorific value.
B) The <em>calorific value</em> for a fuel is the amount of heat - measured in Joules- which is produced during the complete combustion of the fuel. It is expressed in Joules per Kg of fuel (J/kg).
C) From the data:
mass of fuel = 2 kg
heat produced = 48,000 KJ
We calculate the calorific value by dividing the heat produced by the mass of fuel, as follows:
calorific value = heat produced/mass of fuel = (48,000 KJ)/(2 kg)= 24,000 kJ/kg
Since 1 KJ= 1000 J, we can express the calorific value in J/kg as follows:
24,000 kJ/kg x 1000 J/1 kJ = 2.4 x 10⁷ J/kg
Gravity when the mantle cools it becomes heavier and therfore sinks
Answer: 1mole of NaCl
Explanation:
Looking at the given equation :CaCl2 (aq) + Na2CO3 (aq) → CaCO3 (s) + 2 NaCl (aq)
The following deductions can be made,
1 mole of CaCl2 reacted with 1 mole of Na2CO3 to produce 1mole of CaCO3 and 2NaCl.
If 0.500 mole of CaCO3 was produced from this same reaction,it therefore means that we multiply the coefficients of both reactants and product by 0.500 to know the moles of NaCl that will also come out from this same reaction..
So we have the new equation to be:
0.5CaCl2 (aq) + 0.5 Na2CO3 (aq) → 0.5 CaCO3 (s) + 1 NaCl (aq)
From the new equation,1mole of NaCl will be produce alongside 0.5mole of CaCO3,when 0.5mole of CaCl2 reacts with.0.5mole Na2CO3.
