The temperature of the wind as that decreases the volume and the pressure of the balloon to the given values is 14.09°C.
<h3>What is Combined gas law?</h3>
Combined gas law put together both Boyle's Law, Charles's Law, and Gay-Lussac's Law. It states that "the ratio of the product of volume and pressure and the absolute temperature of a gas is equal to a constant.
It is expressed as;
P₁V₁/T₁ = P₂V₂/T₂
Given the data in the question;
- Initial volume V₁ = 14.5L
- Initial pressure P₁ = 0.980atm
- Initial temperature T₁ = 20.0°C = 293.15K
- Final pressure P₂ = 740.mmHg = 0.973684atm
We substitute our given values into the expression above.
P₁V₁/T₁ = P₂V₂/T₂
( 0.980atm × 14.5L )/293.15K = ( 0.973684atm × 14.3L )/T₂
14.21Latm / 293.15K = 13.92368Latm / T₂
14.21Latm × T₂ = 13.92368Latm × 293.15K
14.21Latm × T₂ = 4081.72679LatmK
T₂ = 4081.72679LatmK / 14.21Latm
T₂ = 287.24K
T₂ = 14.09°C
Therefore, the temperature of the wind as that decreases the volume and the pressure of the balloon to the given values is 14.09°C.
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Answer:
The standard moisture content specification for hard gelatin capsules is between 13 % w/w and 16 % w/w. This value can vary depending upon the conditions to which they are exposed: at low humidity's they will lose moisture and become brittle, and at high humidity's they will gain moisture and soften.
Explanation:
Hope this helps!
They may break bonds,form new compounds, new ions etc...
Answer:
ΔH° = -186.2 kJ
Explanation:
Hello,
This case in which the Hess method is applied to compute the required chemical reaction. Thus, we should arrange the given first two reactions as:
(1) it is changed as:
SnCl2(s) --> Sn(s) + Cl2(g)...... ΔH° = 325.1 kJ
That is why the enthalpy of reaction sign is inverted.
(2) remains the same:
Sn(s) + 2Cl2(g) --> SnCl4(l)......ΔH° = -511.3 kJ
Therefore, by adding them, we obtain the requested chemical reaction:
(3) SnCl2(s) + Cl2(g) --> SnCl4(l)
For which the enthalpy change is:
ΔH° = 325.1 kJ - 511.3 kJ
ΔH° = -186.2 kJ
Best regards.
Answer:
3.5 × 10⁵ g of salt
Explanation:
<em>What is the mass (grams) of salt in 10.0 m³ of ocean water?</em>
We have this data:
- 1.000 mol salt is equal to 58.44 g salt
- 1.0 L of ocean water contains 0.60 mol of salt
We will need the following relations:
We can use proportions:
