A student determines that 23.1 J of heat are required to raise the temperature of 6.67 g of an
Answer:
Both require time, but velocity requires displacement and speed requires distance
Explanation:
For calculating speed we require time and distance because speed is defined as the distance per unit time and as speed is a scalar quantity it does not have any direction
But for calculating the velocity we require time as well as displacement because velocity is defined as the displacement per unit time and as velocity is a vector quantity it has direction
Displacement is the shortest distance between the initial position and the final position and it has a specified direction as well
I have not read the article, and so I don't know in what context this is referring to. But I do know that <span>Infra-sound is made up of a really low frequency sound, beyond the range of hearing by humans. </span>Helpful? :)
Answer:
189.71 secs
Explanation:
We know that decomposition is a first order reaction;
So;
ln[A] = ln[A]o - kt
But;
[A]o = 1.00 M
[A] = 0.250 M
t =135 s
Hence;
ln[A] - ln[A]o = kt
k = ln[A] - ln[A]o/t
k = ln(1) - ln(0.250)/135
k =0 - (-1.386)/135
k = 1.386/135
k= 0.01
So time taken now will be;
ln[A] - ln[A]o = kt
t = ln[A] - ln[A]o/k
t = ln (3) - ln(0.450)/0.01
t = 1.0986 - (-0.7985)/0.01
t = 1.0986 + 0.7985/0.01
t = 189.71 secs
Answer:
1.208x10⁻³M and 392.5ppm La(NO3)3
Explanation:
The reaction that occurs is:
La2O3 + 6HNO3 → 2La(NO3)3 + 3H2O
Molarity is defined as the moles of solute (In this case, LaO3) per liter of solution. And ppm, are mg of solute per liter of solution.
To solve this question we must find the moles of La(NO3)3 produced and its mass in milligrams to find molarity and ppm:
<em>Moles La2O3 -Molar mass: 325.81g/mol-</em>
0.1968g * (1mol / 325.81g) = 6.04x10⁻⁴ moles La2O3
<em>Moles La(NO3)3:</em>
6.04x10⁻⁴ moles La2O3 * (2mol La(NO3)3 / 1mol La2O3) = 1.208x10⁻³ moles La(NO3)3
<em>Molarity:</em>
1.208x10⁻³ moles La(NO3)3 / 1L =
<h3>1.208x10⁻³M</h3>
<em>Mass La(NO3)3 -Molar mass: 324.92g/mol-</em>
1.208x10⁻³ moles La(NO3)3 * (324.92g / mol) = 0.392.5g La(NO3)3
In mg:
392.5mg La(NO3)3 / 1L =
392.5ppm La(NO3)3
