Answer:
225,563,910 m/s
Explanation:
speed of light in water = speed of light in the air/refractive index
speed of light in water = 3.00 x 10^8 m/s/1.33
speed of light in water = 225,563,910 m/s
Answer:
527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
When the heat added or removed from a substance causes a change in temperature in it, this heat is called sensible heat.
In other words, the sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a change in temperature without there being a change in physical state (solid, liquid or gaseous). The equation that allows to calculate this heat exchange is:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT=Tfinal-Tinitial is the change in temperature.
In this case:
- c= 4.184

- m=21 g
- ΔT=Tfinal-Tinitial=28 °C - 34 °C=-6 °C
Replacing:
Q= 4.184
* 21 g* (-6 C)
Q= - 527.184 J
To lower the temperature, heat has to be given, for that the final temperature must be lower than the initial temperature; and it receives the name of transferred heat and has a negative value, as in this case.
<u><em>
527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.</em></u>
Answer:
1) pure water
2) 0.75 m CaCl2
3) 1.0 m NaCl
4) 0.5 m KBr
5) 1.5 m glucose (C6H12O6)
Explanation:
Boiling point elevation is a colligative property. Coligative properties are properties that depend on the amount of solute present in the system. The boiling point of solvents increase due to the presence of solutes.
The boiling point elevation depends on the number of particles the solute forms in solution and the molality of the solute. The more the number of particles formed by the solute and the greater the molality of the solute, the greater the magnitude of boiling point elevation.
The order of decreasing hoping point elevation is;
1) 0.75 m CaCl2
2) 1.0 m NaCl
3) 0.5 m KBr
4) 1.5 m glucose (C6H12O6)
The answer is 60.3% magnesium, 39.7% oxygen.
Solution:
The chemical equation for the reaction is 2 Mg + O2 → 2 MgO.
Since magnesium reacts completely with oxygen, it is the limiting reactant in the reaction. Hence, we can use the number of moles of magnesium to get the mass of MgO produced:
moles of magnesium = 14.7g / 24.305g mol-1
= 0.6048 mol
mass of MgO = 0.6048mol Mg(2 mol MgO/2mol Mg)(40.3044g MgO/1 mol MgO)
= 24.376g MgO
We can now solve for the percentage of magnesium:
% Mg = (14.7g Mg / 24.376g MgO)*100% = 60.3%
We also use the number of moles of magnesium to get the mass of oxygen consumed in the reaction:
mass of O2 = 0.6048 mol Mg (1mol O2 / 2mol Mg) (31.998g / 1mol O2)
= 9.676g
The percentage of oxygen is therefore
% O2 = (9.676g O2 / 24.376g MgO)*100%
= 39.7%
Notice that we can just subtract the magnesium's percentage from 100% to get
% O2 = 100% - 60.3% = 39.7%
Answer:
is b
Explanation:
es la b porque estpy estudiando lo mismo