I think the best answer would be D. Dolphin!
Answer:
14.57g
Explanation:
Given parameters:
Mass of dish + ball = 15.6g
Initial volume of water in the cylinder = 26.7mL
Final volume of water in the cylinder = 38.9mL
Mass of dish = ?
Unknown
Mass of the ball = ?
Solution;
Since the mass of ball and dish is 15.6g,
Mass of the ball =Mass of ball + dish - mass of the dish
Insert the parameters and solve;
Mass of the ball = 15.6g - 1.03g = 14.57g
<h3>
Answer:</h3>
The total concentration of ions in a 0.75 M solution of HCl is 1.5 M
That is; 0.75 M H⁺ and 0.75 M Cl⁻
<h3>
Explanation:</h3>
- Concentration or molarity is the number of moles of a compound or an ion contained in one liter of solution. It is measured in moles per liter (M).
- The concentration of ions making a compound is determined by the ratio of moles of the compound and the constituents ions.
- For instance, HCl dissociates to give H⁺ and Cl⁻
HCl(aq) → H⁺(aq) + Cl⁻(aq)
- Therefore, since the mole ratio between HCl and the constituent ions H⁺ and Cl⁻ is 1:1, then 0.75 M of HCl dissociates to give 0.75 M H⁺ and 0.75 m Cl⁻
- Hence the total concentration of ions in a 0.75 M solution of HCl is 1.5 M (0.75 M H⁺ and 0.75 M Cl⁻)
<em>Gasoline</em><em> </em><em>in</em><em> </em><em>water</em><em> </em><em>=</em><em> </em><em>insoluble</em><em> </em>
<em>acetone</em><em> </em><em>in</em><em> </em><em>nail</em><em> </em><em>polished</em><em> </em><em>=</em><em> </em><em>soluble</em><em> </em>
<em>salt</em><em> </em><em>in</em><em> </em><em>alcohol</em><em> </em><em>=</em><em> </em><em>soluble</em><em> </em>
<em>oil</em><em> </em><em>in</em><em> </em><em>vinegar</em><em> </em><em>=</em><em> </em><em>insoluble</em><em> </em>
<em>tawas</em><em> </em><em>in</em><em> </em><em>water</em><em> </em><em>=</em><em> </em><em>soluble</em><em>.</em><em>.</em><em>.</em>
<em>Sorry</em><em> </em><em>if</em><em> </em><em>i</em><em> </em><em>am</em><em> </em><em>incorrect</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
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>