All categories

3 years ago

Why are both accuracy and precision important in making a measurement

1 answer:

4 0

It is because precise amount measured is required for the reaction to take place chemically, and an accurate conclusion will prevent any dangerous and misleading conclusions made due to inaccurate data.

For example, in physics, every measurement we state the instrument uncertainty after the measurement, stating that it is not a definite certain measurement, but the smaller the value for the uncertainty, the more precise that instrument's data is.

Precise is important in terms of knowing how many moles or grams of reactant is required for this reaction to complete without any shortage or excess chemically. Accuracy is based on the conclusions we make in regards to the data and observations we make experimentally.

You might be interested in

Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp

Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

$\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}$

For calculating edge length,

$a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}$

For calculating $M_{ave}$ , we use the formula

$M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}$

Similarly for calculating $(\rho)_{ave}$ , we use the formula

$\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}$

From the periodic table, masses of the two elements can be written

$M_{Fe}= 55.85g/mol$

$M_{V}=50.941g/mol$

Specific density of both the elements are

$(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3$

Putting $M_{ave}$ and $\rho_{ave}$ formula's in edge length formula, we get

$a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}} \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}} \right )} \right ]^{1/3}$

$a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}} \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}} \right )} \right ]^{1/3}$

By calculating, we get

$a=2.89\times10^{-8}cm=0.289nm$

7 0

3 years ago

What decomposers are in a estuary ?

belka [17]

They are Common Soil Bacteria Nematodes.

Hope this helps:)

7 0

4 years ago

Which of the following has the electron configuration [ar] 4s23d5 ?

nignag [31]

Answer:

Mn Manganese

Explanation:

5 0

3 years ago

Elements and compounds are more similar to each other than they are to mixtures because:

Dovator [93]

Answer:

elements and compounds can only be liquids or solids, but mixtures can be solids, liquids or gases

Explanation:

Element: A substance that is made up of only one type of atom. Compound: A substance that is made up of more than one type of atom bonded together. Mixture: A combination of two or more elements or compounds which have not reacted to bond together; each part in the mixture retains its own properties.

7 0

3 years ago

A company desires to produce nickel from refining nickel oxide and sulfide ores. the first step of the process will produce a ni

inn [45]

Answer is: increase the concentration of NiCO₄ and withdraw CO from the products as it is forming.

Balanced chemical reaction: NiCO₄(g) ⇄ Ni(s) + 4CO(g).
According to Le Chatelier's Principle the position of equilibrium moves to counteract the change:
1) the position of equilibrium will move to the right, so that the concentration of nickel carbonyl gas decreases again and produce more nickel.
2) the position of equilibrium will move to the right, if we decrease concentration of carbon monoxide, more nickel carbonyl gas will decompose and produce more nickel.

4 0

3 years ago