I think the answer is false
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
A) SrBr2
b) C3P4
c) Be3(PO4)2
dCa3P2
e) MnS2O3
d) NO
Given:
n = 0.0021 mol
M = 44 gram/mol
Required: m or the mass of the molecule
Solution:
M = m/n
Rearranging the expression,
m = M x n
m = 44 g/mol x 0.0021 mol
Evaluating the expression and cancelling units, we obtain,
m = 0.0924 grams
M is the molar mass. This value is very useful in chemistry calculations especially in calculating the amounts of reactants and products for a chemical reaction.
Answer:
The final temperature is:- 7428571463.57 °C
Explanation:
The expression for the calculation of heat is shown below as:-
Where,
is the heat absorbed/released
m is the mass
C is the specific heat capacity
is the temperature change
Thus, given that:-
Mass of water = 1.75 mg = 0.00175 g ( 1 g = 0.001 mg)
Specific heat of water = 4.18 J/g°C
Initial temperature = 35 °C
Final temperature = x °C
![\Delta T=(x-35)\ ^0C/tex] Q = [tex]1.3\times 10^4](https://tex.z-dn.net/?f=%5CDelta%20T%3D%28x-35%29%5C%20%5E0C%2Ftex%5D%0A%3C%2Fp%3E%3Cp%3EQ%20%3D%20%5Btex%5D1.3%5Ctimes%2010%5E4)
kcal
Also, 1 kcal = 4.18 kJ = 
J
So, Q = 
J = 54340000 J
So,
Thus, the final temperature is:- 7428571463.57 °C
