Answer:
36.09% is the mass percent composition of calcium in calcium chloride
Explanation:
Mass percent is defined as one hundred times the ratio between the mass of a compound (In this case, Calcium), and the total mass of the sample:
<em>Mass Percent = Mass compound / Total mass of the sample × 100</em>
<em />
Computing the values of the problem:
Mass Percent = 0.690g / 1.912g * 100
Mass percent =
<h3>36.09% is the mass percent composition of calcium in calcium chloride</h3>
Answer:
1.6 grams
Explanation:
We need to prepare 100 mL (0.100 L) of a 0.10 M CuSO₄ solution. The required moles of CuSO₄ are:
0.100 L × 0.10 mol/L = 0.010 mol
The molar mass of CuSO₄ is 159.61 g/mol. The mass corresponding to 0.010 moles is:
0.010 mol × (159.61 g/mol) = 1.6 g
We should use 1.6 grams of CuSO₄.
Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
![K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D%7D)
Given:
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
![0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}](https://tex.z-dn.net/?f=0.011%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B0.10%7D)
So,
![\left[I_{Equilibrium} \right]^2=0.011\times 0.10](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.011%5Ctimes%200.10)
![\left[I_{Equilibrium} \right]^2=0.0011](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.0011)
![\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%3D3.3166%5Ctimes%2010%5E%7B-2%7D%5C%20M)
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>
All of the above are chemical reactions.
If that is a choice
<span>Scientists ignore the forces of attraction between particles in a gas under ordinary conditions</span><span> because the particles in a gas are apart and moving fast, rather than clustered and moving slow, therefore the forces of attraction are too weak to have a visible effect.</span>