<span>Henry divides 1.060 g by 1.0 mL to find the density of his water sample.
</span>He should include THREE significant figures in the density value that hereports.
Answer:
None of these are correct, because there is no way to balance this equation, but I hope these steps help you figure out your answer.
Explanation:
Count out the single amounts of elements you have on both sides of the equation. To be balanced, you need to have the exact same for each element.
Before balanced Left side.
Cl-2
O-8
H-2
Before balanced right side.
H-1
Cl-1
O-3
That means we need to increase Hydrogen, Chlorine and Oxygen on the right for sure and see how that affects the equation. You can keep adding the Coefficients until the # of elements begin to match on each side.
(I tried to balance this equation, it doesn't work, there is too much on the reactants side for what the product is.)
We use the osmotic pressure to determine the concentration of the solute in the solution. Then, we multiply the volume of the solution to determine the number of moles of solute particles. We need to establish to equations since we have two unknowns, the mass of of each solute. We do as follows:
osmotic pressure = CRT
<span>C = 7.75 / 0.08205 (296.15) = 0.3189 mol / L</span>
<span>moles of particles = C*V = 0.3189*0.250 =0.0797 mol </span>
<span>0.0797 = moles of sucrose + 2*moles of salt </span>
<span>x + 2y = 0.0797 </span>
<span>and </span>
<span>x(MMsucrose) + y(MMNaCl) = 10.2</span>
<span>342x + 58.5y = 10.2
</span>
<span>solve for x and y
</span>
<span>x = 0.0252 mol sucrose</span>
<span>y = 0.0273 mol NaCl
</span>
<span>mass Sucrose = 0.0252(342) = 8.6184 g </span>
<span>mass NaCl = 0.0273(58.5) = 1.5971 g </span>
<span>% NaCl = (1.5971 / 10.2)*100 = 15.66%</span>
When dissolved in water, acids donate hydrogen ions (H+). Hydrogen ions are hydrogen atoms that have lost an electron and now have just a proton, giving them a positive electrical charge. ... If a solution has a high concentration of H+ ions, then it is acidic.
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.
