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Tpy6a [65]
1 year ago
8

A student measures the mass of 40 mL of a liquid 3 times. They obtain mass readings of 28.5 g,

Chemistry
1 answer:
algol [13]1 year ago
4 0

It can be said that the measurements of the students are precise but not accurate.

<h3>What is accuracy and precision?</h3>

Accuracy is the degree of conformity of a measure to a true or standard value while precision is the ability of a measurement to be reproduced consistently.

Simply put, measurements are said to be accurate when they are close to a true or standard value but are said to be precise when they are close to one another.

According to this question, a student measures the mass of 40 mL of a liquid 3 times and obtained the following mass readings: 28.5g, 28.4g and 28.2g.

Based on the measurement, the measured data are precise because they are close to one another but are not accurate because they are not near the standard value of 37.9g.

Learn more about precision and accuracy at: brainly.com/question/15276983

#SPJ1

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2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s
Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

6 0
3 years ago
If you assume this reaction is driven to completion because of the large excess of one ion, what is the concentration of [Fe(SCN
viktelen [127]

Answer : The concentration of [Fe(SCN)]^{2+} is, 4.32\times 10^{-4}M

Explanation :

When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is SCN^- and Fe^{3+} is excess reagent.

First we have to calculate the moles of KSCN.

\text{Moles of }KSCN=\text{Concentration of }KSCN\times \text{Volume of solution}

\text{Moles of }KSCN=0.00180M\times 0.006L=1.08\times 10^{-5}mol

Moles of KSCN = Moles of K^+ = Moles of SCN^- = 1.08\times 10^{-5}mol

Now we have to calculate the concentration of [Fe(SCN)]^{2+}

\text{Concentration of }[Fe(SCN)]^{2+}=\frac{\text{Moles of }[Fe(SCN)]^{2+}}{\text{Volume of solution}}

Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L

\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M

Thus, the concentration of [Fe(SCN)]^{2+} is, 4.32\times 10^{-4}M

7 0
3 years ago
15 mL of acid 2 M was added to 20 mL of base 2 M into a calorimeter at room temperature (24 oC). The reaction mixture reached a
erik [133]

Answer:

here:

Explanation:

The changes in temperature caused by a reaction, combined with the values of the specific heat and the mass of the reacting system, makes it possible to determine the heat of reaction.

Heat energy can be measured by observing how the temperature of a known mass of water (or other substance) changes when heat is added or removed. This is basically how most heats of reaction are determined. The reaction is carried out in some insulated container, where the heat absorbed or evolved by the reaction causes the temperature of the contents to change. This temperature change is measured and the amount of heat that caused the change is calculated by multiplying the temperature change by the heat capacity of the system.

The apparatus used to measure the temperature change for a reacting system is called a calorimeter (that is, a calorie meter). The science of using such a device and the data obtained with it is called calorimetry. The design of a calorimeter is not standard and different calorimeters are used for the amount of precision required. One very simple design used in many general chemistry labs is the styrofoam "coffee cup" calorimeter, which usually consists of two nested styrofoam cups.

When a reaction occurs at constant pressure inside a Styrofoam coffee-cup calorimeter, the enthalpy change involves heat, and little heat is lost to the lab (or gained from it). If the reaction evolves heat, for example, very nearly all of it stays inside the calorimeter, the amount of heat absorbed or evolved by the reaction is calculated.

8 0
3 years ago
¡cual es el % masa/volumen de una mezcla que posee 40 gramos de colorante en 5 litros de mezcla?
Firlakuza [10]

Answer:

what

Explanation:

7 0
3 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
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