Answer:
![\boxed {\boxed {\sf 1.87 \J/g \textdegree C}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%201.87%20%5CJ%2Fg%20%5Ctextdegree%20C%7D%7D)
Explanation:
We are asked to find the specific heat capacity of a liquid. We are given the heat added, the mass, and the change in temperature, so we will use the following formula.
![q= mc\Delta T](https://tex.z-dn.net/?f=q%3D%20mc%5CDelta%20T)
The heat added (q) is 47.1 Joules. The mass (m) of the liquid is 14.0 grams. The specific heat (c) is unknown. The change in temperature (ΔT) is 1.80 °C.
- q= 47.1 J
- m= 14.0 g
- ΔT= 1.80 °C
Substitute these values into the formula.
![47.1 \ J = (14.0 \ g) * c * (1.80 \textdegree C)](https://tex.z-dn.net/?f=47.1%20%5C%20J%20%3D%20%2814.0%20%5C%20g%29%20%2A%20c%20%2A%20%281.80%20%5Ctextdegree%20C%29)
Multiply the 2 numbers in parentheses on the right side of the equation.
![47.1 \ J = (14.0 \ g * 1.80 \textdegree C)*c](https://tex.z-dn.net/?f=47.1%20%5C%20J%20%3D%20%2814.0%20%5C%20g%20%2A%201.80%20%5Ctextdegree%20C%29%2Ac)
![47.1 \ J = (25.2 \ g*\textdegree C) *c](https://tex.z-dn.net/?f=47.1%20%5C%20J%20%3D%20%2825.2%20%5C%20g%2A%5Ctextdegree%20C%29%20%2Ac)
We are solving for the heat capacity of the liquid, so we must isolate the variable c. It is being multiplied by 25.2 grams * degrees Celsius. The inverse operation of multiplication is division, so we divide both sides of the equation by (25.2 g * °C).
![\frac {47.1 \ J}{(25.2 g *\textdegree C)} = \frac {(25.2 g *\textdegree C)*c}{{(25.2 g *\textdegree C)}}](https://tex.z-dn.net/?f=%5Cfrac%20%7B47.1%20%5C%20J%7D%7B%2825.2%20g%20%2A%5Ctextdegree%20C%29%7D%20%3D%20%5Cfrac%20%7B%2825.2%20g%20%2A%5Ctextdegree%20C%29%2Ac%7D%7B%7B%2825.2%20g%20%2A%5Ctextdegree%20C%29%7D%7D)
![\frac {47.1 \ J}{(25.2 g *\textdegree C)} =c](https://tex.z-dn.net/?f=%5Cfrac%20%7B47.1%20%5C%20J%7D%7B%2825.2%20g%20%2A%5Ctextdegree%20C%29%7D%20%3Dc)
![1.869047619 \ J/g *\textdegree C = c](https://tex.z-dn.net/?f=1.869047619%20%5C%20J%2Fg%20%2A%5Ctextdegree%20C%20%3D%20c)
The original measurements of heat, mass, and temperature all have 3 significant figures, so our answer must have the same. For the number we found that is the hundredth place. The 9 in the thousandth place to the right tells us to round the 6 up to a 7.
![1.87 \ J/ g * \textdegree C =c](https://tex.z-dn.net/?f=1.87%20%5C%20J%2F%20g%20%2A%20%5Ctextdegree%20C%20%3Dc)
The heat capacity of the liquid is approximately 1.87 J/g°C.
Answer:
The electric force between them decreases
Explanation:
The force between two charged particle is given by :
![F=\dfrac{kq_1q_2}{r^2}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7Bkq_1q_2%7D%7Br%5E2%7D)
Where
r is the distance between charges
If the distance between the charges is increased, the electric force gets decreased as there is an inverse relation between force and distance.
Hence, the correct option is (c) "The electric force between them decreases"
The δs∘rxn for the reaction
→
will be -146 J/K.
Entropy would be a measurable physical characteristic and a scientific notion that is frequently connected to a condition of disorder, unpredictability, or uncertainty.
Entropy would be a measurement of the system's unpredictability or disorder. The entropy increases as randomness do. It has broad properties as well as a state function. It has the unit
.
Entropy of the reaction can be calculated by the reaction.
Δ
= 2 mol ×
×
- 1 mol × ![S^{0} (O_{2} )](https://tex.z-dn.net/?f=S%5E%7B0%7D%20%28O_%7B2%7D%20%29)
Δ
= 2 mol × 240 J/mol.K - 2 mol × 210 J/mol.K-1 mol ×205.2 J/mol.K
Δ
= -146.8 J/K
Therefore, the δs∘rxn for the reaction
→
will be -146 J/K.
To know more about reaction
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Answer:
27.6mL of LiOH 0.250M
Explanation:
The reaction of lithium hydroxide (LiOH) with chlorous acid (HClO₂) is:
LiOH + HClO₂ → LiClO₂ + H₂O
<em>That means, 1 mole of hydroxide reacts per mole of acid</em>
Moles of 20.0 mL = 0.0200L of 0.345M chlorous acid are:
0.0200L ₓ (0.345mol / L) = <em>6.90x10⁻³ moles of HClO₂</em>
To neutralize this acid, you need to add the same number of moles of LiOH, that is 6.90x10⁻³ moles. As the LiOH contains 0.250 moles / L:
6.90x10⁻³ moles ₓ (1L / 0.250mol) = 0.0276L of LiOH =
<h3>27.6mL of LiOH 0.250M</h3>