Answer
Manual samplers are prone to effects of temperature, speed of wind and air concentrations.
Explanation
Manual samplers face several challenges that can act as drawbacks to obtaining accurate results. They are subjected to effects of sampling duration where long sampling times are needed to obtain adequate mass for detection. Manual samplers face challenges when measuring non-volatile species because particles are observed into the adsorption medium at a slower rate of diffusion.
<span>HCl<span>(aq)</span>+NaOH<span>(aq)</span>→NaCl<span>(aq)</span>+<span>H2</span>O<span>(l)</span></span>
As you can see here, one mole of acid neutralizes one mole of base.
We use the concentration equation, which states that,
<span>c=<span>nv</span></span>
<span>
<span>
<span>
n is the number of moles
</span>
<span>
v is the volume of solution
</span>
</span>
</span>
Rearranging for moles, we get,
<span>n=c⋅v</span>
So, we have:
<span><span>n<span>NaOH</span></span>=0.1 M⋅0.05 L</span>
<span>=0.005 mol</span>
Since one mole of acid neutralizes one mole of base, then we must have: <span><span>n<span>HCl</span></span>=<span>n<span>NaOH</span></span></span>.
And so,
<span><span>c<span>HCl</span></span>=<span><span>n<span>HCl</span></span><span>v<span>HCl</span></span></span></span>
<span>=<span><span>0.005 mol</span><span>0.03 L</span></span></span>
<span>≈0.17 <span>M</span></span>
Answer:
Explanation:
In this case, we have a dilution problem. We have to remember that in the dilution procedure we go from a solution with higher concentration to a solution with lesser concentration. Therefore we have to start with the dilution equation:
Now we can identify the variables:
If we plug all the values into the equation:
And we solve for 
:
I hope it helps!
To calculate number of moles, all you do is divide the given mass by the molecular molar mass:
<span>i.e. 125g / 18g = 6.94444g </span>
<span>Therefore, your answer is (a) 6.94 g</span>
Answer:
when water changes from a liquid on earths surface to a gas in the atmosphere,this is known as evaporation
