Rust does not have the properties to catch onto flames. However, if you light it on fire, then it would probably catch in flames but not instantly and will not continue to burn unless you have soaked it in oil or flammable object or substance. :) Hope this helps!
<span>We can use the heat
equation,
Q = mcΔT </span>
<span>
Where Q is the amount of energy transferred (J), m is
the mass of the substance (kg), c is the specific heat (J g</span>⁻¹ °C⁻<span>¹) and ΔT is the temperature
difference (°C).</span>
According to the given data,
Q = 300 J
m = 267 g
<span>
c = ?
ΔT = 12 °C</span>
By applying the
formula,
<span>300 J = 267 g x c x
12 °C
c = 0.0936 J g</span>⁻¹ °C⁻<span>¹
Hence, specific heat of the given substance is </span>0.0936 J g⁻¹ °C⁻¹.
Answer:
It heats up.
Explanation:
Because thermal energy is based off the temperature of an object and the average kinetic energy of its particles increases. When the average kinetic energy of its particles increases, the object's thermal energy increases. Therefore, the thermal energy of an object increases as its temperature increase.
Answer:
At equilibrium:
[H2] = 0.005 M
[Br2] = 0.105 M
[HBr] = 0.189 M
Explanation:
H2(g) + Br2(g) ⇄ 2HBr
an "x" value will be used from reactant to produced "2x"
so at equilibrium:
[H2] = 0.1 - x
[Br2] = 0.2 - x
[HBr] = 2x
we know that Kc=[HBr]²/[H2][Br2]
Thus 62.5 = (2x)²/(0.1-x)(0.2-x)
this generate a quadratic equation: 58.5x² - 18.75x + 1.25 = 0
the x₁ = 0.23 x₂ = 0.09457
we pick 0.09457 because the two reactants can not make more than what they have. x₁ is higher than both initial reactant concentration
Then we substitute the "x₂" value at equilibrium:
[H2] = 0.1-0.09457 = 0.005 M
[Br2] = 0.2-0.09457 = 0.105 M
[HBr] = 2*0.09457 = 0.189 M
Yes because condensed milk and evaporated milk are similar to one another. However, there won’t be the same sweet flavor but the texture is the same.
