His distance and displacement are the same, which was 400 m
<h3>Further explanation</h3>
Given
Distance = 400 m
time = 2 min
Required
Distance and displacement
Solution
Distance is a scalar quantity that indicates the length of the trajectory that is traveled by an object within a certain interval. Distance has no direction, only has magnitude
Can be simplified distance = totals traveled
Displacement is a vector quantity that shows changes in the position of objects in a certain interval of time. Displacement has magnitude and direction
Can be simplified displacement = distanced traveled from starting point to ending point
From the definition above shows that the displacement and the distance that he traveled have the same value (magnitude), which is equal to 400 m
The value of the two will be different if he starts and finishes at the same point, then the displacement value is zero while the distance he has traveled is still 0
Answer:
The temperature for ![\Delta G^o=0[/tex is [tex]T=328.6 K](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D0%5B%2Ftex%20is%20%5Btex%5DT%3D328.6%20K)
Explanation:
The three thermodinamic properties (enthalpy, entropy and Gibbs's energy) are linked in the following formula:
Where:
is Gibbs's energy in kJ
is the enthalpy in kJ
is the entropy in kJ/K
is the temperature in K
Solving:
For 
:
Answer:
Option C. hydroxide ions (OH-).
Explanation:
A base is a substance which dissolves in water to produce hydroxide ion (OH-) as the only negative ion. It therefore means that a base contains more hydroxide ions (OH-).
Answer:
The specific rotation of D is 11.60° mL/g dm
Explanation:
Given that:
The path length (l) = 1 dm
Observed rotation (∝) = + 0.27°
Molarity = 0.175 M
Molar mass = 133.0 g/mol
Concentration in (g/mL) = 0.175 mol/L × 133.0 g/mol
Concentration in (g/mL) = 23.275 g/L
Since 1 L = 1000 mL
Concentration in (g/mL) = 0.023275 g/mL
The specific rotation [∝] = ∝/(1×c)
= 0.27°/( 1 dm × 0.023275 g/mL
)
= 11.60° mL/g dm
Thus, the specific rotation of D is 11.60° mL/g dm
Answer:
Explanation:
Hello,
In this case, we can consider that the given heat of combustion is indeed the heat of reaction since it corresponds to the combustion of propane, which is computed by using the heat formation of all the involved species as shown below:
Thus, since the heat of formation of gaseous carbon dioxide is -393.5 kJ/mol, water -241.8 kJ/mol and oxygen 0 kJ/mol, the heat of formation of propane is:
Best regards.
