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3 years ago

A reproducible measurement is an accurate one. Sometimes true? Always true? Never true?

1 answer:

8 0

Never true.

Accuracy is how close your answer is to the one you were supposed to get.

Precision is how repeatable the answer you got was.

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Calculate the number of joules of heat energy needed to increase the temperature of 25.0 g of metal from 21.0 ºC to 80.0 ºC. The

Harman [31]

Answer:

Q = 768.47 J

Explanation:

Given that,

Mass of the metal, m = 25 g

Initial temperature, T₁ = 21.0 ºC

Final temperature, T₂ = 80.0 ºC

The specific heat of the metal is 0.521 J/gºC.

We know that the heat released due to the change in temperature is given by :

$Q=mc\Delta T\\\\=25\times 0.521\times (80-21)\\Q=768.47\ J$

Hence, 768.47 J of heat energy will be needed.

7 0

3 years ago

Convert 4.992×1010 g to each of the following units. B. Mg c. mg D. metric tons

SVETLANKA909090 [29]

I think it is letter D

3 0

4 years ago

A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som

jek_recluse [69]

The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of $-6.4^oC$ and a molal freezing point depression constant $K_f=3.96^oC.kg/mol$ . A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at $-13.6^oC$ . Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

Answer: The mass of glycine that can be dissolved is $1.3\times 10^2g$

Explanation:

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m$

$\text{Freezing point of pure solvent}=\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Freezing point of pure solvent = $-6.4^oC$

Freezing point of solution = $-13.6^oC$

i = Vant Hoff factor = 1 (for non-electrolytes)

$K_f$ = freezing point depression constant = $3.96^oC/m$

$m_{solute}$ = Given mass of solute (glycine) = ?

$M_{solute}$ = Molar mass of solute (glycine) = 75.07 g/mol

$w_{solvent}$ = Mass of solvent = 950. g

Putting values in equation 1, we get:

$-6.4-(-13.6)=1\times 3.96\times \frac{m_{solute}\times 1000}{75.07\times 950}\\\\m_{solute}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\m_{solute}=129.66g=1.3\times 10^2g$