Answer:
The correct answer is 25 mL graduated cylinder (it should be used in all the cases)
Explanation:
In order to measure 25.00 ml sample of a solution it should be used a 25 mL graduated cylinder, as it is previously and properly calibrated. The other laboratory glassware, beaker and erlenmeyer, have graduations which are approximate, so they are used when exact volumes are not needed.
ii) graduated cylinder has the least uncertainly. It is more accurate than a beaker or erlenmeyer (to within 1%)
iii) A 25 mL graduated cylinder should be used because it is the most accurate lab glassware (between those were mentioned: beaker, erlenmeyer).
<u>Answer:</u> The concentration of
in three significant figures will be 0.899 mol/L.
<u>Explanation:</u>
For the given reaction:

The above reaction follows zero order kinetics. The rate law equation for zero order follows:
![k=\frac{1}{t}([A_o]-[A])](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%28%5BA_o%5D-%5BA%5D%29)
where,
k = rate constant for the reaction = 
t = time taken = 10 sec
= initial concentration of the reactant = 0.962 mol/L
[A] = concentration of reactant after some time = ?
Putting values in above equation, we get:
![6.28\times 10^{-3}=\frac{1}{10}(0.962-[A])](https://tex.z-dn.net/?f=6.28%5Ctimes%2010%5E%7B-3%7D%3D%5Cfrac%7B1%7D%7B10%7D%280.962-%5BA%5D%29)
![[A]=0.899mol/L](https://tex.z-dn.net/?f=%5BA%5D%3D0.899mol%2FL)
Hence, the concentration of
in three significant figures will be 0.899 mol/L.
Answer:
1.5L of NaNO3 must be present
Explanation:
<em>That contains 200g of sodium nitrate. Round to 2 significant digits</em>
To solve this question we need to convert the mass of NaNO3 to moles using its molar mass (85g/mol). With the moles and the molar concentration we can find the volume in liters of the solution:
<em>Moles NaNO3:</em>
200g * (1mol / 85g) = 2.353 moles NaNO3
<em>Volume:</em>
2.353 moles NaNO3 * (1L / 1.60moles) =
<h3>1.5L of NaNO3 must be present</h3>
Try this solution:
if 1 mole is 55.8 gr., then 2.25 moles the mass should be 55.8*2.25=125.55 gr.
Answer: 125.55 gr.
Explanation:
Since the ratio is Cu/Fe, if some Fe were lost due to spillage, the Cu/Fe ratio would INCREASE because Fe would be lower.