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Answer: Approximately
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Work Shown:
We have the following variables
- Q = 714 joules = heat required
- m = 52 grams = mass
- c = specific heat = unknown
= 82-30.5 = 51.5 = change in temperature
note: the symbol is the uppercase Greek letter delta. It represents the difference or change in a value.
Apply those values into the formula below. Solve for c.
The specific heat of the unknown metal is roughly
Answer:
a) 0,5 mol O₂; 1 mol NO₂
b)
Liters of NO = 22,4 L
Liters of O₂ = 11,2 L
Liters of NO₂ = 22,4 L
c) NO = 30g
O₂ = 16g
NO₂ = 46g
d) 7,41g HCl
e) 107,7 g/mol
f) 0,0312 moles of O₂; 0,999g of O₂; 699 mL at STP or 803 mL
ii. 51%
g) 32,6L
Explanation:
a) For the reaction:
2 NO + O₂ → 2 NO₂
For 1 mole of NO there are consumed:
1 mol NO × = <em>0,5 mol of O₂</em>
And produced:
1 mol NO × = <em>1 mol of NO₂</em>
b) By ideal gas law:
V = nRT/P
Where n is moles of each compound; R is gas constant (0,082atmL/molK); T is temperature (273,15 K at state conditions); P is pressure (1 atm at STP) and V is volume in liters. Replacing each moles for each compound:
Liters of NO = 22,4 L
Liters of O₂ = 11,2 L
Liters of NO₂ = 22,4 L
c) The mass of each compound are:
1 mol NO× = <em>30g</em>
0,5 mol O₂× = <em>16g</em>
1 mol NO₂× = <em>46g</em>
d) Using:
n = PV / RT
Moles of 4,55 L of HCl (using the values of P = 1 atm; R = 0,082atmL/molK; T = 273,15K) are:
0,203 moles of HCl. In grams:
0,203 mol HCl× = <em>7,41 g of HCl</em>
e) Using:
δRT/P = MW
Where δ is density in g/L (4,81 g/L); R is gas constant (0,082atmL/molK); T is temperature (273,15K); P is pressure (1 atm)
And MW is molecular mass: <em>107,7 g/mol</em>
f) For the reaction:
2 KClO₃ → 2 KCl + 3 O₂
2,550 g of KClO₃ are:
2,550 g of KClO₃× = 0,0208 moles of KClO₃
When these moles reacts completely produce:
0,0208 moles of KClO₃× = <em>0,0312 moles of O₂</em>
In grams:
0,0312 moles of O₂ × = <em>0,999g of O₂</em>
V = nRT/P
At STP, n = 0,0312 mol; R = 0,082atmL/molK;T= 273,15K; P = 1atm; <em>V = 0,699L ≡ 699mL</em>
At 29 °C (302,15K) and 732 torr (0,963 atm)
<em>V = 0,803L ≡ 803mL</em>
ii. 182 mL ≡ 0,182L of O₂ are:
n = PV/RT
moles of O₂ are 7,07x10⁻³. Moles of KClO₃ are:
7,07x10⁻³ moles of O₂× = 0,0106 mol KClO₃. In grams:
0,0106 moles of KClO₃× = 1,300 g of KClO₃.
Thus, percent by mass of KClO₃ in the mixture is:
1,300g/2,550g ×100 = <em>51%</em>
g. Combined gas law says that:
Where:
P₁ = 755 torr; V₁ = 35,9L; T₁ = 26°C (299,15 K); P₂ = 760 torr (STP): T₂ = 273,15K (STP) <em>V₂ = 32,6 L</em>
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I hope it helps!