Answer:
Explanation:
Cubic decimeter is the same unit as liter; so, mole per cubic decimeter is mole per liter, and that is the unit of concentration of molarity. Thus, what is asked is the molarity of the solution. This is how you find it.
1. <u>Take a basis</u>: 1 dm³ = 1 liter = 1,000 ml
2. <u>Calculate the mass of 1 lite</u>r (1,000 ml) of solution:
- density = mass / volume ⇒ mass = density × volume
Here, the density is given through the specific gravity
Scpecific gravity = density of acid / density of water
Take density of water as 1.00 g/ml.
- density of solution = 1.25 g/ml
- mass solution = 1.25 g/ml × 1,000 ml = 1,250 g
3. <u>Calculate the mass of solute</u> (pure acid)
- % m/m = (mass of solute / mass of solution) × 100
- 56 = mass of solute / 1,250 g × 100
- mass of solute = 56 × 1,250g / 100 = 700 g
4. <u>Calculate the number of moles of solute</u>:
- moles = mass in grams / molar mass = 700 g / 70 g/mol = 10 mol
5. <u>Calculate molarity (mol / dm³)</u>
- M = number of moles of solute / liter of solution = 10 mol / 1 liter = 10 mol/liter.
<span> 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH.
</span>
<span>31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH.
</span>
<span>After the reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:
</span>
<span>pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) </span>
<span>= -log(1.8x10^-5*5.8/12.4) = 5.07</span>
If the liquid is at or above its flash point, the flame spread rate is fast, and the entire pool is engulfed within seconds. ... As the liquid temperature decreases, flame radiation must both heat the liquid to the flash point temperature and supply the heat of vaporization.
Answer:
0.252 mol
Explanation:
<em>Given the following reaction: </em>
<em>Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂</em>
<em>How many moles of Ag will be produced from 16.0 g Cu, assuming AgNO3 is available in excess.</em>
First, we write the balanced equation.
Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂
We can establish the following relations.
- The molar mass of Cu is 63.55 g/mol.
- The molar ratio of Cu to Ag is 1:1.
The moles of Ag produced from 16.0 g of Cu are:
Stoichiometry time! Remember to look at the equation for your molar ratios in other problems.
31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag 6851.65
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ → ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ = 107.9 g Ag
∅ | 63.5 g Cu | 1 mol Cu | 1 mol Ag 63.5
There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.
