Answer:
Explanation:
Q = Ce . m .ΔT
Q : calor
Ce : calor especifico
m: masa
ΔT : variación de temperatura
capacidad térmica : 0,550 cal / °C
lectura :
por 1 °C se tiene 0,550 cal
por lo tanto tenemos datos de la temperatura y del calor
pero no olvidar las unidades en el sistema internacional :
Ce : J / kg . K
J: joules
kg: kilogramo
K: kelvin
pasar de gramos a kilogramos
pasar de calorías a joules
pasar de grado celsius a kelvin
1000g equivale a 1kg
15g equivale a 0,015 kg
K= °C + 273 ⇒ formula para pasar de grado celsius a kelvin
K= 1 + 273
K= 274
1 caloría equivale a 4,184 joules
0,550 caloría equivale a 2,3012 joules
ahora como todos los datos ya están en el S.I remplazamos en la formula
Q = Ce . m .ΔT
2,3012 = Ce . 0,015.274
Ce=0,5599 J / kg. K
Answer:
- Dalton used creativity to modify Proust's experiment and interpret the results.
- Thomson used creativity to interpret the results of the cathode ray tube experiment.
Explanation:
give me brainlest
Volume in liters of Carbon dioxide : 0.672
<h3>Further explanation</h3>
Reaction(combustion of butane-C₄H₁₀)
2C₄H₁₀+13O₂⇒8CO₂+10H₂O
mol butane (MW=58,12 g/mol) :
mol CO₂ : mol C₄H₁₀ = 8 : 4, so mol CO₂ :
At STP, 1 mol = 22.4 L, so volume CO₂ :
The question is incomplete. The complete question is stated below:
Two point charges are held at the corners of a rectangle as shown in the figure. The lengths of sides of the rectangle are 0.050 m and 0.150 m. Assume that the electric potential is defined to be zero at infinity.
a. Determine the electric potential at corner A.
b. What is the electric potential energy of a +3 µC charge placed at corner A?
Answer / Explanation:
a )V(A) = 1 / 4πe° ( - 5 5x10∧6C / 0.150m + 2x10∧6C / 0.050m )
The answer to the equation above is : = +6.0x10∧4 j/c
b) U(A) = qV(A)= (3.0x10∧6C) (6.0x10∧4 . j/c) =
The answer to the equation above is : =0.18 J
Explanation:
Where V(A) is equivalent to the electric potential
U(A) is equivalent to the electric potential energy
<span> a) AgNO3 (aq) + KCl (aq) --> AgCl (s) + KNO3 (aq) this is balanced chemical equation
</span><span>AgCl is the precipitate so we shall not split it
</span><span>Ag+(aq) + NO3 -(aq) + K+ (aq) + Cl-(aq) --> AgCl (s) + K+(aq) + NO3-(aq)
</span><span>Ag+ (aq) + Cl- (aq) ---> AgCl (s) after removing ions</span>
