Answer:
The specific heat of the mineral is 0.1272J/g°C
Explanation:
The sample is given energy to the calorimeter and the sample of water.
The energy released for the sample is equal to the energy absorbed for both the calorimeter and the water:
C(Sample)*m*ΔT = C(Calorimeter)*ΔT + C(water)*m*ΔT
<em>Where C is specific heat</em>
<em>m is mass of the sample and water</em>
<em>And ΔT is change in temperature</em>
<em />
C(Sample)*149g*(92.7°C-23.7°C) = 12.8J/K*(23.7°C-20.0°C) + 4.184J/g°C*81.4g*(23.7°C-20.0°C)
C(Sample)*10281g°C = 47.36J + 1260.1J
C(Sample) = 0.1272J/g°C
<h3>The specific heat of the mineral is 0.1272J/g°C</h3>
<em />
Molecular weight of P (Phosphorous) = 30.97g/mol
Hydrogen is just 1 g/mol.
How many moles is 34g of PH3?
Get the weight of PH3 (30.97 + (3X1)) = 33.97g/mol
So 34g/33.97g/mol = 1.0009 moles.
I bet for this problem it's easier to round this to 1.
If you look at just the moles in the equation:
P4(s) + 6 H2(g) → 4 PH3(g)
OR
1 + 6 → 4
If 1 (P4) gives us 4 (PH3), what gives us 1 (PH3)?
1/4 = x/1
solve for x
Answer:
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Answer:
8. littoral I hope it helps