Answer:
0.1127 grams of Mg is used to produce one gram of silver
Explanation:
Write a balanced equation
Mg(s) +2Ag + (aq) --> 
(aq) + 2Ag(s)
The atomic mass of magnesium is 24.31 g/mol and the atomic mass of silver is 107.84g/mol
1 g Ag of silver is produced from [(1 mol Mg)(24.31 g/mol)] /[ (2mol Ag)(107.84g/mol)]
1 g Ag of silver is produced from 0.1127 grams of Mg
0.1127 grams of Mg is used to produce one gram of silver
Answer:here is the answer
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Explanation:
Answer:
9.28moles
Explanation:
Given parameters:
volume = 11.1L
pressure = 204atm
temperature = 24°C = 24 + 273 = 297K
Unknown:
Number of moles of air in the cylinder = ?
Solution:
To solve this problem, we apply the ideal gas equation;
PV = nRT
P is the pressure
V is the volume
n is the number of moles
R is the gas constant = 0.082atmdm³mol⁻¹K⁻¹
T is the temperature
Now insert the parameters and find n;
204 x 11.1 = n x 0.082 x 297
226.4 = 24.4n
n = 9.28moles
Answer:
ΔH = -55.92 kJ
Explanation:
<u>Step 1:</u> Data given
1 mol NaOH and 1 mol HBr initially at 22.5 °C are mixed in 100g of water
After mixing the temperature rises to 83 °C
Specific heat of the solution = 4.184 J/g °C
Molar mass of NaOH = 40 G/mol
Molar mass of HBr = 80.9 g/mol
<u>Step 2: </u>The balanced equation
NaOH + HBr → Na+(aq) + Br-(aq) + H2O(l)
<u>Step 3:</u> mass of NaOH
Mass = moles * Molar mass
Mass NaOH = 1 * 40 g/mol
Mass NaOH = 40 grams
Step 4: Mass of HBr
Mass HBr = 1 mol * 80.9 g/mol
Mass HBr = 80.9 grams
Step 5: Calculate ΔH
ΔH = m*c*ΔT
ΔH= (100 + 40 + 80.9) * 4.184 * (83-22.5)
ΔH= 220.9 * 4.184 * 60.5
ΔH= 55916.86 J = 55.92 kJ
Since this is an exothermic reaction, the change in enthalpy is negative.
ΔH = -55.92 kJ
