Use equation density = mass/volume
Known/Given
d= 10.5g/cm^3
v= 24.0 cm^3
Use that equation i just mentioned...
Solution
Rewrite the formula as the forumla for solving density into mass
Which is... mass = 10.5g/cm^3 × 24.0cm^3 = 252g
So thats the answer
252g
Hope i could help
Explanation:
As the Earth rotates on its axis and revolves around the Sun, several different effects are produced. When the new moon comes between the Earth and the Sun along the ecliptic, a solar eclipse is produced. When the Earth comes between the full moon and the Sun along the ecliptic, a lunar eclipse occurs.
Answer:
The equilbrium constant is 179.6
Explanation:
To solve this question we can use the equation:
ΔG = -RTlnK
<em>Where ΔG is Gibbs free energy = 12.86kJ/mol</em>
<em>R is gas constant = 8.314x10⁻³kJ/molK</em>
<em>T is absolute temperature = 298K</em>
<em>And K is equilibrium constant.</em>
Replacing:
12.86kJ/mol = -8.314x10⁻³kJ/molK*298K lnK
5.19 = lnK
e^5.19 = K
179.6 = K
<h3>The equilbrium constant is 179.6</h3>
Explanation:
Molar mass of 
= 39.1 + 35.5 + 3(16.0) = 122.6 g
Molar mass of KCl = 39.1 + 35.5 = 74.6 g
Molar mass of 
= 32.0 g
According to the equation, 2 moles of reacts to give 3 moles of oxygen.
Therefore, 2 (122.6) = 245.2 g of will give 3 (32.0) = 96.0 g of oxygen. Thus, 245.2 g of gives 96.0 g of oxygen.
(a) Calculate the amount of oxygen given by 2.72 g of as follows.
of
(b) Calculate the amount of oxygen given by 0.361 g of as follows.
of
c) Calculate the amount of oxygen given by 83.6 kg as follows.
of
Convert kg into grams as follows.
= 32731 g of
(d) Calculate the amount of oxygen given by 22.5 mg of as follows.
Convert mg into grams as follows.
of
Answer:
![K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNOBr%5D%5E4_%7Beq%7D%7D%7B%5BNO%5D%5E4_%7Beq%7D%5BBr%5D%5E2_%7Beq%7D%7D)
Explanation:
Hello,
In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:
The suitable equilibrium constant turns out:
Or in terms of the initial equilibrium constant:
Since the second reaction is a doubled version of the first one.
Best regards.
