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3 years ago

How do interpolation and extrapolation differ, which is more reliable

Chemistry

2 answers:

miv72 [106K]3 years ago

7 0

Answer:

These 2 are techniques whereby an individual tries to analyze and observe additional data points or information that may not be recorded and or obvious. From a graph, usually linear.

Interpolation involves the individual to be aware of the pattern observed or relationship between the 2 variables and tries to obtain a value that is in between 2 recorded or defined data points.

Extrapolation involves the individual to estimate and imagine what the graph would appear as if further data was collected, based on the previous recordings. This can be found out due to the relationship between the 2 variables in the graph.

Explanation:

MAXImum [283]3 years ago

6 0

These 2 are techniques whereby an individual tries to analyze and observe additional data points or information that may not be recorded and or obvious. From a graph, usually linear.

Interpolation involves the individual to be aware of the pattern observed or relationship between the 2 variables and tries to obtain a value that is in between 2 recorded or defined data points.

Extrapolation involves the individual to estimate and imagine what the graph would appear as if further data was collected, based on the previous recordings. This can be found out due to the relationship between the 2 variables in the graph.

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A 15.00g solid mixture containing Ca(OH)2, among other non-basic components, was neutralized with 0.2000g of HCl. What was the m

azamat

Answer:

1.373 wt% Ca(OH)₂

Explanation:

Sample mix = 15.0g

Ca(OH)₂(aq) + 2HCl(aq) => CaCl₂(aq) + 2H₂O(l)

moles HCl = 0.2000g / 36 g·mol⁻¹ 0.0056 mol

moles Ca(OH)₂ = 1/2(moles HCl) = 1/2(0.0056 mol) = 0.0028 mol

mass Ca(OH)₂ = 0.0028 mol ( 74 g/mol ) = 0.206 g

mass % Ca(OH)₂ = (0.206/15.0)100% = 1.373 wt%

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3 years ago

If you have 5.22*1022 molecules of glucose (C6H12O6) how many grams is this?

lianna [129]

Answer:

5.22*1022 molecules of glucose (C6H12O6) is equal to 15.08 grams

Explanation:

Number of molecule of glucose in one mole $= 6.023 * 10^{23}$

Now mass of one mole of glucose $= 180$ grams

Number of moles in $6.023 * 10^{23}$ molecules of glucose $= 1$

Number of moles in $1$ molecules of glucose = $\frac{1}{6.023 * 10^{23}}$

Number of moles in $5.22 * 10^{22}$ molecules of glucose = $\frac{5.22* 10^{22}}{6.023 * 10^{23}}$ = $0.083$ moles

Weight of $0.083$ moles

$= 0.083 * 180\\= 15.08$ grams

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3 years ago

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4 years ago

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3 years ago

If 0.25 mol of br2 and 0.55 mol of cl2 are introduced into a 3.0-l container at 400 k, what will be the equilibrium concentratio

Vladimir79 [104]

Answer is: the equilibrium concentration of Br₂ is 0,02 mol/L.
Chemical reaction: Br₂ + Cl₂ → 2BrCl.
Kc = 7,0.
c₀(Br₂) = 0,25 mol ÷ 3 L.
c₀(Br₂) = 0,083 mol/L.
c₀(Cl₂) = 0,55 mol ÷ 3 L.
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Kc = c(BrCl)² ÷ c(Br₂) · c(Cl₂).
7 = (2x)² ÷ (0,083 mol/L - x) · (0,183 mol/L - x).
7 = 4x² ÷ (0,083 mol/L - x) · (0,183 mol/L - x).
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3 years ago

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