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: negative acellaration or mass.
Explanation:
the first reason why is that i got that quistion right. and when objects are unbalanced it gives negative acellaration
It is 2.1 x 10^3 because your base number needs to be in between 1 and 10, and the number you are converting is non-decimal, so the exponent is positive. It is 10^3 because you are moving the decimal 3 places to the right
Answer:
[CaSO₄] = 36.26×10⁻² mol/L
Explanation:
Molarity (M) → mol/L → moles of solute in 1L of solution
Let's convert the volume from mL to L
250 mL . 1L/1000 mL = 0.250L
We need to determine the moles of solute. (mass / molar mass)
12.34 g / 136.13 g/mol = 0.0906 mol
M → 0.0906 mol / 0.250L = 36.26×10⁻² mol/L
Answer:
21.16 MPa
Explanation:
Partial pressure of oxygen = 5.62 MPa
Total gas pressure = 26.78 MPa
But
Total pressure of the gas= sum of partial pressures of all the constituent gases in the system.
This implies that;
Total pressure of the system = partial pressure of nitrogen + partial pressure of oxygen
Hence partial pressure of nitrogen=
Total pressure of the system - partial pressure of oxygen
Therefore;
Partial pressure of nitrogen= 26.78 - 5.62
Partial pressure of nitrogen = 21.16 MPa
