Answer:
We need 1.1 grams of Mg
Explanation:
Step 1: Data given
Volume of water = 78 mL
Initial temperature = 29 °C
Final temperature = 78 °C
The standard heats of formation
−285.8 kJ/mol H2O(l)
−924.54 kJ/mol Mg(OH)2(s)
Step 2: The equation
The heat is produced by the following reaction:
Mg(s)+2H2O(l)→Mg(OH)2(s)+H2(g)
Step 3: Calculate the mass of Mg needed
Using the standard heats of formation:
−285.8 kJ/mol H2O(l)
−924.54 kJ/mol Mg(OH)2(s)
Mg(s) + 2 H2O(l) → Mg(OH)2(s) + H2(g)
−924.54 kJ − (2 * −285.8 kJ) = −352.94 kJ/mol Mg
(4.184 J/g·°C) * (78 g) * (78 - 29)°C = 15991.248 J required
(15991.248 J) / (352940 J/mol Mg) * (24.3 g Mg/mol) = 1.1 g Mg
We need 1.1 grams of Mg
Answer: I think the answer is 2.5 M.
Explanation: Formula: M = n/v
M= 10 mol / 4L
M= 2.5 M
Answer:
Number of molecules = 1.8267×10^20
Explanation:
From the question, we can deuced that the gases behave ideally, the we can make use of the ideal gas equation, which is expressed below;
PV = nRT
where
P =pressure
V =volume
n = the number of moles
R is the gas constant equal to 0.0821 L·atm/mol·K
T is the absolute temperature
Given:
P = 6.75 atm;
T = 290.0 k,
; V = 1.07 cm³ = 0.001 L
( 6.75 atm)(0.00107 L) = n(0.0821 L·atm/mol·K)(290K)
n = 3.0335167*10^-4 moles
But there are 6.022×10²³ molecules in 1 mole,
Number of molecules = 1.8267×10^20
Answer:
I think it's B but I could be wrong so really sorry if I am
