Answer: The balloons volume would increase.
Explanation: This is because when the balloon enters a warmer climate it expands due to more freedom within the molecules movement.
The answer is going to be “chemical bond” hope you have a good day and hope this helps
Answer:
Explanation:
The solution contain 0.01 M concentration of Ba²⁺
0.01M concentration of Ca²⁺
Ksp ( solubility constant) for BaSO₄ = 1.07 × 10⁻¹⁰
Ksp for CaSO₄ = 7.10 × 10⁻⁵
(BaSO₄) = (Ba²⁺) (SO₄²⁻)
1.07 × 10⁻¹⁰ = 0.01 M (SO₄²⁻)
1.07 × 10⁻¹⁰ / 0.01 = ( SO₄²⁻)
1.07 × 10⁻⁸ M = ( SO₄²⁻)
so the minimum of concentration of concentration sulfate needed is 1.07 × 10⁻⁸ M
For CaSO₄
CaSO₄ = ( Ca²⁺) ( SO₄²⁻)
7.10 × 10⁻⁵ = 0.01 (SO₄²⁻)
(SO₄²⁻) = 7.10 × 10⁻⁵ / 0.01 = 7.10 × 10⁻³ M
so BaSO₄ will precipitate first since its cation (0.01 M Ba²⁺) required a less concentration of SO₄²⁻ (1.07 × 10⁻⁸ M ) compared to CaSO₄
b) The minimum concentration of SO₄²⁻ that will trigger the precipitation of the cation ( 0.01 M Ba²⁺) that precipitates first is 1.07 × 10⁻⁸ M
Answer:
D +405.0kJ mol-¹
Explanation:
Since bond energy is the energy required to break a bond, the energy of dissociation of X₂H₆ = +2775 kJmol⁻¹.
Since there is one X-X bond and six X-H bonds,
Bond energy of one X-X bond + Bond energy of six X-H bonds = energy of dissociation of X₂H₆.
Since bond energy of one X-H bond = 395 kJ mol⁻¹, then
Bond energy of one X-X bond + Bond energy of six X-H bonds = energy of dissociation of X₂H₆
Bond energy of one X-X bond + 6 × one X-H bond = +2775 kJmol⁻¹.
Bond energy of one X-X bond + 6 × 395 kJ mol⁻¹ = +2775 kJmol⁻¹.
Bond energy of one X-X bond + 2370 kJ mol⁻¹ = 2775 kJmol⁻¹
Bond energy of one X-X bond = 2775 kJmol⁻¹ - 2370 kJ mol⁻¹
Bond energy of one X-X bond = +405 kJmol⁻¹
Preparing 15 mg/gl working standard solution from a 20 mg/dl stock solution will require the application of the dilution principle.
Recalling the principle:
initial volume x initial molarity = final volume x final molarity
Since we were not given any volume to work with, we can as well just take an arbitrary volume to be prepared. Let's assume that the stock solution is 10 mL and we want to prepare 15 mg/gl from it:
Applying the dilution principle:
10 x 20 = final volume x 15
final volume = 200/15
= 13.33 mL
This means that in order to prepare 13.33 mL, 15 mg/l working standard solution from 10 ml, 20 mg/dl stock solution, 3.33 mL of the diluent must be added to the stock solution.
More on dilution principle can be found here: brainly.com/question/11493179
