Answer:
Explanation:
Method 1 proportion
1 mole of chromium is 52 grams
11.9 moles = x grams
1/11.9 = 52/x Cross multiply
x = 11.9 * 52
x = 618.8 grams
Now I have used an approximate mass for Chromium. The answer you get here is expected to reflect the weigth given on your periodic table Use that to get your answer. You should give a number very close to mine. Round to 3 places as in 619.
Method Two Formula
mols = given mass / molecular mass
11.9 = given mass / 51.9961 Multiply both sides by 51.9961
11.9 *51.9961 = given mass
given mass = 618.75
given mass = 619
When the little fish die they sink to the bottom, and after time the skeletons build up on top of each other.
Assuming the concentration of stock solution is 50% sodium phosphate buffer solution, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.
<h3>What volume of a stock Sodium phosphate buffer and water is needed to 12 mL of 25% sodium phosphate buffer of pH 4?</h3>
The process of preparing solutions from stock solutions of higher concentration is known as dilution.
Dilution is done with the aid of the dilution formula given below:
where
- C1 is the concentration of stock solution
- V1 is the volume of stock solution required to prepare a diluted solution
- C2 is the concentration of the diluted solution prepared
- V2 is the final volume of the diluted solution
From the data provided:
C1 is not given
V1 is unknown
C2 = 25%
V2 = 12 mL
- Assuming C1 is 50% solution
Volume of stock, V1, required is calculated as follows:
V1 = C2V2/C1
V1 = 25 × 12 /50
V1 = 6 mL
Therefore, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.
Learn more about dilution formula at: brainly.com/question/7208546
Answer:
Water will boil at 
.
Explanation:
According to clausius-clapeyron equation for liquid-vapor equilibrium:
![ln(\frac{P_{2}}{P_{1}})=\frac{-\Delta H_{vap}^{0}}{R}[\frac{1}{T_{2}}-\frac{1}{T_{1}}]](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_%7B2%7D%7D%7BP_%7B1%7D%7D%29%3D%5Cfrac%7B-%5CDelta%20H_%7Bvap%7D%5E%7B0%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D-%5Cfrac%7B1%7D%7BT_%7B1%7D%7D%5D)
where, 
and 
are vapor pressures of liquid at 
(in kelvin) and 
(in kelvin) temperatures respectively.
Here, = 760.0 mm Hg, = 373 K, = 314.0 mm Hg
Plug-in all the given values in the above equation:
![ln(\frac{314.0}{760.0})=\frac{-40.7\times 10^{3}\frac{J}{mol}}{8.314\frac{J}{mol.K}}\times [\frac{1}{T_{2}}-\frac{1}{373K}]](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B314.0%7D%7B760.0%7D%29%3D%5Cfrac%7B-40.7%5Ctimes%2010%5E%7B3%7D%5Cfrac%7BJ%7D%7Bmol%7D%7D%7B8.314%5Cfrac%7BJ%7D%7Bmol.K%7D%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D-%5Cfrac%7B1%7D%7B373K%7D%5D)
or, 
So, 
Hence, at base camp, water will boil at
