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3 years ago

Assesses the consistency of observations by different observers.

1 answer:

3 0

The consistency that should be assessed for observations via different observers is inter-rater reliability.

The following information should be considered:

The inter-rater reliabilities mean the decision consistency that lies between 2 reviewers & the significant validity component.
It should be used for assessing the consistency via various types of observers.

Therefore we can conclude that the consistency that should be assessed for observations via different observers is inter-rater reliability.

Learn more: brainly.com/question/22780158

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BRAINLIEST IF CORRECT

tensa zangetsu [6.8K]

Answer

its physical adaptions

6 0

3 years ago

Read 2 more answers

Which is not a layer found in the earth's mantle? A. Upper mantle B. asthenosphere C. lithosphere D.crust

Debora [2.8K]

Answer:

LITHOSPHERE

Explanation:

5 0

3 years ago

The substances below are listed by increasing specific heat capacity value. Starting at 30 Celsius, they absorb 100 kJ of therma

Georgia [21]

Answer:

Silver.

Explanation:

To obtain the right answer to question, let us calculate the change in temperature for each substance assuming they all have the same mass as 100g.

This is illustrated below:

1. For Siver:

Mass (M) = 100g

Specific heat capacity (C) = 0.239J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 0.239)

ΔT = 4184°C

2. For Aluminium:

Mass (M) = 100g

Specific heat capacity (C) = 0.921J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 0.921)

ΔT = 1086°C

3. For Lithium:

Mass (M) = 100g

Specific heat capacity (C) = 3.56J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 3.56 )

ΔT = 281°C

4. For water:

Mass (M) = 100g

Specific heat capacity (C) = 4.184J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 4.184)

ΔT = 239°C

Summary

Temperature change of each substance is given below

1. Silver => 4184°C

2. Aluminum => 1086°C

3. Lithium => 281°C

4. Water => 239°C

From the calculations made above, Silver has the highest rise in temperature.

4 0

3 years ago

• If 2.5 moles of sugar are added to 2.0 L of a water/lemon juice solution, what is the molarity of the lemonade?

d1i1m1o1n [39]

The answer is 4.5 because you add rhem

6 0

3 years ago

10 m3 of carbon dioxide is originally at a temperature of 50 °C and pressure of 10 kPa. Determine the new density and volume of

Dimas [21]

Answer : The new density and new volume of carbon dioxide gas is 0.2281 g/L and $7.2m^3$ respectively.

Explanation :

First we have to calculate the new or final volume of carbon dioxide gas.

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 10 kPa

$P_2$ = final pressure of gas = 15 kPa

$V_1$ = initial volume of gas = $10m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $50^oC=273+50=323K$

$T_2$ = final temperature of gas = $75^oC=273+75=348K$

Now put all the given values in the above equation, we get:

$\frac{10kPa\times 10m^3}{323K}=\frac{15kPa\times V_2}{348K}$

$V_2=7.2m^3$

The new volume of carbon dioxide gas is $7.2m^3$

Now we have to calculate the new density of carbon dioxide gas.

$PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}$

Formula for new density will be:

$\rho_2=\frac{P_2M}{RT_2}$

where,

$P_2$ = new pressure of gas = 15 kPa

$T_2$ = new temperature of gas = $75^oC=273+75=348K$

M = molar mass of carbon dioxide gas = 44 g/mole

R = gas constant = 8.314 L.kPa/mol.K

$\rho$ = new density

Now put all the given values in the above equation, we get:

$\rho_2=\frac{(15kPa)\times (44g/mole)}{(8.314L.kPa/mol.K)\times (348K)}$

$\rho_2=0.2281g/L$

The new density of carbon dioxide gas is 0.2281 g/L

5 0

3 years ago