The client's lab values are sodium 166 mEq/L, potassium 5.0 mEq/L, chloride 115 mEq/L, and bicarbonate 35 mEq/L. What condition
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Answer:
Lead shows the greatest temperature change upon absorbing 100.0 J of heat.
Explanation:
Q = Energy gained or lost by the substance
m = mass of the substance
c = specific heat of the substance
ΔT = change in temperature
1) 10.0 g of copper
Q = 100.0 J (positive means that heat is gained)
m = 10.0 g
Specific heat of the copper = c = 0.385 J/g°C
2) 10.0 g of aluminium
Q = 100.0 J (positive means that heat is gained)
m = 10.0 g
Specific heat of the aluminium= c = 0.903 J/g°C
3) 10.0 g of ethanol
Q = 100.0 J (positive means that heat is gained)
m = 10.0 g
Specific heat of the ethanol= c = 2.42 J/g°C
4) 10.0 g of water
Q = 100.0 J (positive means that heat is gained)
m = 10.0 g
Specific heat of the water = c = 4.18J/g°C
5) 10.0 g of lead
Q = 100.0 J (positive means that heat is gained)
m = 10.0 g
Specific heat of the lead= c = 0.128 J/g°C
Lead shows the greatest temperature change upon absorbing 100.0 J of heat.
Answer:
103.9 g
Explanation:
- C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
First <u>we convert 54.0 g of propane (C₃H₈) into moles</u>, using its <em>molar mass</em>:
- 54.0 g ÷ 44 g/mol = 1.23 mol C₃H₈
Then we <u>convert 1.23 moles of C₃H₈ into moles of CO₂</u>, using the <em>stoichiometric coefficients</em>:
- 1.23 mol C₃H₈ *
= 3.69 mol CO₂
We <u>convert 3.69 moles of CO₂ into grams</u>, using its <em>molar mass</em>:
- 3.69 mol CO₂ * 44 g/mol = 162.36 g
And <u>apply the given yield</u>:
- 162.36 g * 64.0/100 = 103.9 g