The gram formula mass is Molar mass. The mass of 1.0 moles is :
3) 48.0 g
Answer:
1.95g of Mg(OH)2 are needed
Explanation:
Mg(OH)2 reacts with HCl as follows:
Mg(OH)2 + 2 HCl → MgCl2 + 2H2O
<em>Where 1 mole of Mg(OH)2 reacts with 2 moles of HCl</em>
To solve this question we must find the moles of acid. Then, with the chemical equation we can find the moles of Mg(OH)2 and its mass:
<em>Moles HCl:</em>
158mL = 0.158L * (0.106mol / L) = 0.01675 moles HCl
<em>Moles Mg(OH)2:</em>
0.01675 moles HCl * (2mol Mg(OH)2 / 1mol HCl) = 0.3350 moles Mg(OH)2
<em>Mass Mg(OH)2 -Molar mass: 58.3197g/mol-</em>
0.3350 moles Mg(OH)2 * (58.3197g / mol) =
<h3>1.95g of Mg(OH)2 are needed</h3>
Answer:
Explanation:
a) In an exothermic reaction, the energy transferred to the surroundings from forming new bonds is ___more____ than the energy needed to break existing bonds.
b) In an endothermic reaction, the energy transferred to the surroundings from forming new bonds is ___less____ than the energy needed to break existing bonds.
c) The energy change of an exothermic reaction has a _____negative_______ sign.
d) The energy change of an endothermic reaction has a ____positive________ sign.
The energy changes occur during the bonds formation and bonds breaking.
There are two types of reaction endothermic and exothermic reaction.
Endothermic reactions:
The type of reactions in which energy is absorbed are called endothermic reactions.
In this type of reaction energy needed to break the bond are higher than the energy released during bond formation.
For example:
C + H₂O → CO + H₂
ΔH = +131 kj/mol
it can be written as,
C + H₂O + 131 kj/mol → CO + H₂
Exothermic reaction:
The type of reactions in which energy is released are called exothermic reactions.
In this type of reaction energy needed to break the bonds are less than the energy released during the bond formation.
For example:
Chemical equation:
C + O₂ → CO₂
ΔH = -393 Kj/mol
it can be written as,
C + O₂ → CO₂ + 393 Kj/mol
Polyethylene terephthalate commonly abbreviated as PET
Answer:
V₂ = 1070 mL or 1.07 L
Solution:
Data Given;
P₁ = 1170 mmHg
V₁ = 915 mL
T₁ = 24 °C + 273 K = 297 K
P₂ = 842 mmHg
V₂ = ?
T₂ = - 23 °C + 273 K = 250 K
According to Ideal gas equation,
P₁ V₁ / T₁ = P₂ V₂ / T₂
Solving for V₂,
V₂ = P₁ V₁ T₂ / P₂ T₁
Putting Values,
V₂ = (1170 mmHg × 915 mL × 250 K) ÷ (842 mmHg × 297 K)
V₂ = 1070 mL or 1.07 L
