Your answer is C. Both gasoline and litter would need to be physically separated from the water, because neither bonds with the water.
0.370 mol metal oxide = 55.45 g
<span>1 mol = 55.45/0.370 = 149.86 g </span>
<span>in 1 mol there are 3 mol O = 16 * 3 = 48 g of O </span>
<span>there is 48/149.86 * 100% O in the sample </span>
<span>the sample has 48/149.86 * 0.370 = 0.119 g O</span>
Answer:
the nucleus is called the mass number and, expressed in amu, is approximately equal to the mass of the atom
Explanation:
<span>4NH</span>₃<span> + 6NO → 5N</span>₂<span> + 6H</span>₂<span>O
mol of NO = </span>
=
= 0.93 mol
Based on the balance equation mole ratio of NH₃ : NO is 4 : 6
= 2 : 3
If mol of NO = 0.93 mol
then mol of NH₃ =
= 0.62 mol
Mass of ammonia = mol × molar mass
= 0.62 mol × 17.03 g/mol
= 10.54 g
Therefore B is the best answer
Answer:
The new equilibrium concentration of HI: <u>[HI] = 3.589 M</u>
Explanation:
Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M
Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M
<em>Given chemical reaction:</em> H₂(g) + I₂(g) → 2 HI(g)
The equilibrium constant (
) for the given chemical reaction, is given by the equation:
![K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%20%7B%5BHI%5D%5E%7B2%7D%7D%7B%5BH_%7B2%7D%5D%5C%3A%20%5BI_%7B2%7D%5D%7D)
<u><em>At the original equilibrium state:</em></u>

<u><em>Therefore, at the new equilibrium state:</em></u>
![\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88](https://tex.z-dn.net/?f=%5CRightarrow%20%5BHI%5D%5E%7B2%7D%20%3D%20713.59%20%5Ctimes%200.01805%20%3D%2012.88)
![\Rightarrow [HI] = \sqrt {12.88} = 3.589 M](https://tex.z-dn.net/?f=%5CRightarrow%20%5BHI%5D%20%3D%20%5Csqrt%20%7B12.88%7D%20%3D%203.589%20M)
<u>Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M</u>