Answer:
the protons
Explanation:
the number of protons is constant for each sample of the element. If the number of protons differs, then it is a different element.
The specific heat capacity of the metal given the data from the question is 0.66 J/gºC
<h3>Data obtained from the question</h3>
- Mass of metal (M) = 76 g
- Temperature of metal (T) = 96 °C
- Mass of water (Mᵥᵥ) = 120 g
- Temperature of water (Tᵥᵥ) = 24.5 °C
- Equilibrium temperature (Tₑ) = 31 °C
- Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC
- Specific heat capacity of metal (C) =?
<h3>How to determine the specific heat capacity of the metal</h3>
The specific heat capacity of the sample of the metal can be obtained as follow:
Heat loss = Heat gain
MC(M –Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)
76 × C × (96 – 31) = 120 × 4.184 × (31 – 24.5)
C × 4940 = 3263.52
Divide both side by 4940
C = 3263.52 / 4940
C = 0.66 J/gºC
Learn more about heat transfer:
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Answer:
84 g
Explanation:
The molecular weight of nitrogen is 14.01 g/mol. That of oxygen is 16.00 g/mol, so the weights of the various elements available are ...
From NO
N: (14.01/(14.01+16.00))·(80 g) = 37.35 g
O: 80 g -37.35 g = 42.65 g
From O₂
O: 16 g
__
In each mole of NO₂, the weight of the oxygen is 2(16.00) = 32 g. The weight of the nitrogen is 1(14.01) = 14.01 g.
From the available oxygen, we can produce ...
(42.65 g +16 g)/(32.00 g/mol) = 1.83 mol of NO₂
From the available nitrogen, we can produce ...
(37.35 g)/(14.01 g/mol) = 2.67 mol of NO₂
Clearly, <em>the reaction is limited by the amount of available oxygen</em>. Then the mass of the NO₂ that can be produced is ...
(1.83 mol)(32.00 +14.010 g/mol = 84.3 g ≈ 84 g.
84 grams of NO₂ are produced.