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4 years ago

How can estimating measurements in an experiment lead to inaccurate results?

1 answer:

4 0

Estimating measurements in an experiment leads to inaccurate results because they were guessed. Estimation is roughly guessing the results therefore it leads you to a wrong answer. By guessing you're assuming the answer, you don't know if the answer is correct and you don't know if that answer is the actual result. Say if you were estimating something in your science lab such as how much the battery has drained in a certain area, well using this you don't actually know if that's true and if someone were to use your results they may end up having a problem sense you ESTIMATED IT.

Sorry if this answer isn't sweet and simple but more explained, it's kind of hard explaining this. I hope this helped!

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A crystallographer measures the horizontal spacing between molecules in a crystal. The spacing is

liubo4ka [24]

The spacing is 14.4 nm (nanometers).
It is 14.4 * 10^(-9) m
And 1 mm = 1 * 10^(-3) m
14.4 nm * 105 = 1,512 nm = 1.512 micrometers = 0.001512 mm
Or : 1.512 * 10^(-3) mm
Answer: The total width of crystal is 0.001512 mm.

3 0

3 years ago

50.0g of N2O4 is introduced into an evacuated 2.00L vessel and allowed to come to equilibrium with its decomposition product,N2O

Lady_Fox [76]

The mass of N2O4 in the final equilibrium mixture is 39.45 grams.

The decomposition reaction of N2O4 to 2NO2 can be expressed as:

$\mathbf{N_2O_4{(g)} \leftrightarrow 2NO_{2(g)}}$

From the parameters given:

The mass of N2O4 = 50.0 g
The molar mass of N2O4 = 92.011 g/mol

The number of mole of N2O4 can be determined as:

$\mathbf{Moles \ of \ N_2O_4 = \dfrac{50.0 g}{92.011 g/mol}}$

$\mathbf{Moles \ of \ N_2O_4 = 0.5434 \ moles}$

The volume of the vessel in which N2O4 was evacuated is = 2.0 L

From stochiometry, the concentration of $\mathbf{[N2O4] = \dfrac{mole \ of \ N_2O_4}{volume \ of \ N_2O_4}}$

$\mathbf{[N2O4] = \dfrac{0.5434}{2}}$

$\mathbf{[N2O4] = 0.2717 \ M}$

The I.C.E table can be computed as:

$\mathbf{N_2O_4{(g)}}$ ↔ $\mathbf{ 2NO_{2(g)}}$

Initial 0.2717 0

Change -x +2x

Equilibrium (0.2717 - x) 2x

The equilibrium constant from the I.C.E table can be expressed as:

$\mathbf{K_c = \dfrac{[NO]^2}{[N_2O_4]}}$

$\mathbf{K_c = \dfrac{(2x)^2}{(0.2717-x)}}$

Recall that; Kc = 0.133

∴

$\mathbf{0.133 = \dfrac{4x^2}{(0.2717-x)}}$

0.0361 - 0.133x = 4x²

4x² + 0.133x - 0.0361 = 0

By solving the above quadratic equation, we have;

x = 0.07978

The Concentration of [NO2] = 2x

[NO2] = 2 (0.07978)
[NO2] = 0.15956 M

The Concentration of [N2O4] = 0.2717 - x

[N2O4] = 0.2717 - 0.07978
[NO2] = 0.19192 M

Again, from the decomposition reaction, we can assert that;

$\mathbf{N_2O_4{(g)} \leftrightarrow 2NO_{2(g)}}$

0.19192 M of N2O4 decompose to produce 0.15956 M of NO2.

However, if 5.0 g of NO2 is injected into the vessel, then the number of moles of NO2 injected becomes;

$\mathbf{= \dfrac{5.0 \ g}{46 \ g/mol}} \\ \\ \\ \mathbf{= 0.10869\ moles}$

The Molarity of NO2 injected now becomes:

$\mathbf{= \dfrac{00.10869 }{2} } \\ \\ \\ \mathbf{= 0.05434 \ M }$

So, the new moles of [NO2] becomes = 0.15956 + 0.05434

= 0.2139 M

The new I.C.E table can be computed as:

$\mathbf{N_2O_4{(g)}}$ ↔ $\mathbf{ 2NO_{2(g)}}$

Initial 0.19192 0.2139 M

Change +x -2x

Equilibrium (0.19192 + x) (0.2139 -2x)

NOTE: The injection of NO2 makes the reaction proceed in the backward direction.

The equilibrium constant from the I.C.E table can be expressed as:

$\mathbf{K_c = \dfrac{[NO]^2}{[N_2O_4]}}$

$\mathbf{K_c = \dfrac{(0.2139 - 2x)^2}{(0.19192+x)}}$

Recall that; Kc = 0.133

$\mathbf{0.133= \dfrac{(0.2139 - 2x)^2}{(0.19192+x)}}$

By solving for x;

x = 0.2246 or x = 0.0225

We need to consider the value of x that is less than the initial concentration of NO2(0.2139 M) which is:

x = 0.0225

Now, the final concentration of [N2O4] = (0.19192 + 0.0225)M

= 0.21442 M

The final number of moles of N2O4 = Molarity(concentration) × volume

The final number of moles of N2O4 = (0.21442 × 2) moles

The final number of moles of N2O4 = 0.42884 moles

The mass of N2O4 in the final equilibrium mixture is:

= final number of moles × molar mass of N2O4

= 0.42884 moles × 92 g/mol

= 39.45 grams

Learn more about the decomposition of N2O4 here:

brainly.com/question/25025725

6 0

3 years ago

Which statement is true about the elements in the periodic table?

aleksandr82 [10.1K]

The correct answer is elements in family 7 have similar properties because elements in the same group or family have the same number of valence electrons.

4 0

3 years ago

Read 2 more answers

The electronegativity is 2.1 for h and 1.9 for pb. based on these electronegativities pbh4 would be expected to

Sedbober [7]

PbH4 will be formed as a result of a polar covalent bond between the H and the Pb.

Since H is more electronegative than the Pb, it is, thus, expected that the H would be able to pull the electron charge towards itself. This will result in the H being negative.

Based on this:
PbH4 would be expected to have polar covalent bonds with a partial negative charges on the H atoms.

8 0

3 years ago

If there are 1,000 mg in 1 gram. Then how many mg are in 2.5 grams?

mihalych1998 [28]

Answer:

1 gram =1000mg

Explanation:

Then 2.5g=2.5*1000

2500mg

7 0

3 years ago

How can estimating measurements in an experiment lead to inaccurate results? ​

How can estimating measurements in an experiment lead to inaccurate results?