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DedPeter [7]
3 years ago
14

Pipettes are used when measuring the volumes of liquids with high degree of precision. The following volumes were obtained durin

g the calibration of a 10.00 mL pipette: 10.15 mL, 9.95 mL, 9.99 mL, and 10.02 mL. Calculate the standard deviation for these measurements.
Chemistry
1 answer:
never [62]3 years ago
6 0

Answer:

0.075

Explanation:

First obtain the mean of the measurement;

Mean = 10.15 + 9.95 + 9.99 + 10.02/4 = 10.03

Then obtain d^2= (mean-score)^2 for each score;

(10.15-10.03)^2 = 0.0144

(9.95-10.03)^2 = 0.0064

(9.99-10.03)^2 = 0.0016

(10.02-10.03)^2= 0.0001

∑d^2= 0.0144 + 0.0064 + 0.0016 + 0.0001

∑d^2= 0.0225

Variance = ∑d^2/N = 0.0225/4 = 0.005625

Standard deviation= √0.005625

Standard deviation= 0.075

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VMariaS [17]

The correct answer is due to rapid conversion of nitrates into nirites in extracellular fluids.

Due to the fact that it is quickly transformed to nitrates and nitrites in the extracellular fluid, nitric oxide (NO) functions as a paracrine signal that only impacts nearby cells. Because it relaxes the smooth muscle cells in blood vessel walls, nitric oxide (NO) causes blood vessels to widen. Cell signaling is a type of cellular communication in which a cell produces a signal to cause changes in neighboring cells, changing the behavior of those cells. Paracrine signaling is one type of cell signaling. Responses to allergens, tissue repair, the development of scar tissue, and blood clotting are a few examples of paracrine signaling. The transmission of signals through synapses between nerve cells is known as paracrine signaling.

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5 0
1 year ago
The rate constant for a certain reaction is measured at two different temperatures:
Talja [164]

Answer: The activation energy Ea for this reaction is 22689.8 J/mol

Explanation:

According to Arrhenius equation with change in temperature, the formula is as follows.

ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]

k_1 = rate constant at temperature T_1 = 2.3\times 10^8

k_2 = rate constant at temperature T_2 = 4.8\times 10^8

E_a= activation energy = ?

R= gas constant = 8.314 J/kmol

T_1 = temperature = 280.0^0C=(273+280)=553K

T_2 = temperature = 376.0^0C=(273+376)=649K

Putting in the values ::

ln \frac{4.8\times 10^8}{2.3\times 10^8} = \frac{-E_{a}}{8.314}[\frac{1}{649} - \frac{1}{553}]

E_a=22689.8J/mol

The activation energy Ea for this reaction is 22689.8 J/mol

3 0
3 years ago
Is sugar ( in water) an example of an element or compound
tresset_1 [31]

Answer:

A compound

Example:

Sugar contains carbon, hydrogen and oxygen. Water (H2O) has two hydrogen atoms (H2) and one oxygen atom (O). So both sugar and water are chemical compounds. Together they are a mix of chemical compounds.

8 0
3 years ago
How many lone pairs of electrons in lewis structure of CH3 OH
Brums [2.3K]
There would be 2 which would be on the oxygen
4 0
2 years ago
9. If 28.56 g of K2O is produced when 25.00 g K is reactedaccording to the following equation, what is the percent yieldof the r
Mumz [18]

Answer

b. 95%

Explanation

Given:

Mass of K₂O produced (actual yield) = 28.56 g

Mass of K that reacted = 25.00 g

Equation: 4K(s) + O₂(g) → 2K₂0(s)

What to find:

The percent yield of K₂O.

Step-by-step solution:

The first step is to calculate the theoretical yield of K₂O produced.

From the balanced equation, 4 mol K produced 2 mol K₂O

Molar mass of K₂O = 94.20 g/mol)

Molar mass of K = 39.10 g/mol)

This means 4 mol x 39.10 g/mol = 156.40 g K produced 2 mol x 94.20 g/mol = 188.40 g K₂O

So 25.00 g K will produce:

\frac{25.00\text{ }g\text{ }K\times188.40\text{ }g\text{ }K₂O}{156.40\text{ }g\text{ }K}=30.1151\text{ }g\text{ }K₂O

Actual yield of K₂O = 28.56 g

Theoretical yield of k₂O = 30.1151 g

The percent yield for the reaction can now be calculated using the formula below:

\begin{gathered} Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\% \\  \\ Percent\text{ }yield=\frac{28.56\text{ }g}{30.1151\text{ }g}\times100\% \\  \\ Percent\text{ }yield=0.9484\times100\% \\  \\ Percent\text{ }yield=94.84\%\approx95\% \end{gathered}

Therefore, the percent yield for the reaction is 95%.

3 0
1 year ago
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