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2 years ago

In data measurement, what does precision refer to?

1 answer:

7 0

Precision refers to the closeness of two or more measurements to each other. Using the example above, if you weigh a given substance five times, and get 3.2 kg each time, then your measurement is very precise. Therefore, your answer is (B)

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What part of an atom has most of its mass

cricket20 [7]

Majority of an atoms mass comes for protons and neutrons which makes up the nucleus.

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2 years ago

_____ energy depends on the motion or position of an object.

Dafna1 [17]

Answer: Mechanical energy depends on the motion or position of an object.

Explanation:

Mechanical energy is the energy that is possessed by an object due to its motion or due to its position. Mechanical energy can be either kinetic energy (energy of motion) or potential energy (stored energy of position).

3 0

3 years ago

Read 2 more answers

A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure

Lesechka [4]

Answer:

the final volume of the gas is $V_2$ = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure $P_{ext}$ :

$P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}$

$P_{ext} = 1.03 \ atm$

The workdone W = $P_{ext}$ V

The change in volume ΔV= $\dfrac{W}{P_{ext}}$

ΔV = $\dfrac{130.1 \ J \times \dfrac{1 \ L \ atm}{ 101.325 \ J} }{1.03 \ atm }$

ΔV = $\dfrac{1.28398717 }{1.03 }$

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial volume = 61.5 mL

The change in volume V is $\Delta V = V_2 -V_1$

$- V_2= - \Delta V -V_1$

multiply through by (-), we have:

$V_2= \Delta V+V_1$

$V_2$ = 1250 mL + 61.5 mL

$V_2$ = 1311.5 mL

∴ the final volume of the gas is $V_2$ = 1311.5 mL

5 0

2 years ago

A 0.2722 g sample of a pure carbonate, X n CO 3 ( s ) , was dissolved in 50.0 mL of 0.1200 M HCl ( aq ) . The excess HCl ( aq )

Alex73 [517]

Answer:

0.00369 moles of HCl react with carbonate.

Explanation:

Number of moles of HCl present initially = $\frac{0.1200}{1000}\times 50.0$ moles = 0.00600 moles

Neutralization reaction (back titration): $NaOH+HCl\rightarrow NaCl+H_{2}O$

According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.

So, excess number of moles of HCl present = number of NaOH added for back titration = $\frac{0.0980}{1000}\times 23.60$ moles = 0.00231 moles

So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles

Hence, 0.00369 moles of HCl react with carbonate.

3 0

3 years ago

Indicate the correct number of sig figs

Dimas [21]

Explanation:

The answer to questions are

A) 4

B) 3

C) 5

D) 3

E) 3

3 0

2 years ago