The specific heat of the metal object with a mass of 22.7g heated to to temperature of 97.0°C and then transferred to an insulated container containing 84.7 g of water at 20.5 ∘C is 0.815J/g°C
How to calculate specific heat?
The specific heat capacity of a metal can be calculated using the calorimetry equation as follows:
Q = mc∆T
Where;
Q = quantity of heat absorbed
m = mass of substance
c = specific heat capacity
∆T = change in temperature
mc∆T (water) = -mc∆T (metal)
84.7 × 4.18 × 3.8 = - (22.7 × c × -72.7)
1345.375 = 1650.29c
c = 0.815J/g°C
Therefore, the specific heat of the metal object with a mass of 22.7g heated to to temperature of 97.0°C and then transferred to an insulated container containing 84.7 g of water at 20.5 ∘C is 0.815J/g°C.
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Answer:
The answer would be a
Explanation:
That is what goes into the plant and oxygen IS an outcome
Hi! I’m pretty sure the answer is B.
Answer:
the sodium nitrate in this investigation is
a) the solute
b) undergoing chemical change
<h3>
Explanation:</h3>
generally, solute is the substance which is added in solvent.
The substance which is taken in relatively more volume is the solvent.
so, here sodium nitrate (NaNO₃) is the solute which is dissolved in water (H₂O) , which is solvent.
The reaction between them is :
NaNO₃ + H₂O ⇒ Na⁺ (aq) + NO₃⁻ (aq) ; (aq) denotes aqueous;
so, here sodium nitrate undergoes chemical change into aqueous sodium ions(Na⁺) and aqueous nitrate ions(NO₃⁻).