Answer:
<em><u>= - 0.38 eV</u></em>
Explanation:
Using Bohr's equation for the energy of an electron in the nth orbital,
E = -13.6 
Where E = energy level in electron volt (eV)
Z = atomic number of atom
n = principal state
Given that n = 6
⇒ E = -13.6 × 
<em><u>= - 0.38 eV</u></em>
<em><u></u></em>
<em>Hope this was helpful.</em>
<em><u></u></em>
Answer:
81°C.
Explanation:
To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat released from water (Q = - 1200 J).
m is the mass of the water (m = 20.0 g).
c is the specific heat capacity of water (c of water = 4.186 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).
∵ Q = m.c.ΔT
∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).
(- 1200 J) = 83.72 final T - 7953.
∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.
<em>So, the right choice is: 81°C.</em>
Answer is "B - 700,000".<span>
<span>Kinetic energy of a single particle (atom or molecule)<span> is directly proportional to its
temperature according to the following equation.</span></span>
KE = (3kT)/2
<span>Where </span>KE<span> is the
kinetic energy of a single atom/molecule (</span>J<span>), </span>k<span> is the Boltzmann
constant (</span>1.381 × 10</span>⁻²³ J/K<span>) and </span>T<span> is the temperature (</span>K<span>) </span><span>
When temperature increases, then the kinetic
energy increases.
<span>If kinetic
energy of atoms increases, then there will be more motions which create many
collisions.</span></span>
Answer:
a) 
b) 1657 €
Explanation:
Hola,
a) En este problema, vamos a considerar el millón de litros de agua anuales, ya que con ellos podemos calcular el calor requerido para dicho calentamiento, sabiendo que la densidad del agua es de 1 kg/L:
Luego, usamos la entalpía de combustión del metano para calcular su requerimiento en kilogramos, sabiendo que la energía ganada por el agua, es perdida por el metano:
b) En este caso, consideramos que a condiciones normales de 1 bar y 273 K, 1 metro cúbico de metano cuesta 0,45 €, con esto, calculamos las moles de metano a dichas condiciones:
Con ello, los kilogramos de metano que cuestan 0,45 €:
Luego, aplicamos la regla de tres:
0.715 kg ⇒ 0.45 €
2630 kg ⇒ X
X = (2630 kg x 0.45 €) / 0.715 kg
X = 1657 €
Regards.
Answer:6.719Litres of Cl2 gas.
Explanation:According to eqn of rxn
2Na +Cl2=2NaCl
P=689torr=689/760=0.91atm
T=39°C+273=312K
according to stoichiometry of the reaction,1Moles of Cl2 gives 2moles of NaCl
But 28g of NaCl was given,we have to convert this to moles by using the relation, n=mass/MW
MW of NaCl=23+35.5=58.5g/mol
n=28g(mass given of NaCl)/58.5
n=0.479moles of NaCl
Going back to the reaction,
if 1moles of Cl2 produces 2moles of NaCl
x moles of Cl2 will give 0.479moles of NaCl.
x=0.479*1/2
x=0.239moles of Cl2.
To find the volume, we use ideal ggas eqn,PV=nRT
V=nRT/P
V=0.239*0.082*312/0.91
V=6.719Litres
