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fomenos
3 years ago
7

What is the purpose of constructing a calibration curve

Chemistry
1 answer:
stellarik [79]3 years ago
5 0

Answer:

Calibration curves are used to understand the instrumental response to an analyte, and to predict the concentration of analyte in a sample.

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a square boat made from iron has overall dimensions of 2.00 cm x 11.0 cm x 11.0 cm. it has a mass of 213 g. water has a density
Stolb23 [73]
Density of boat  =  \frac{MASS}{VOLUME}
                           
                          =  \frac{213 g}{(2 cm  *  11 cm  *  11cm)}
 
                          =   0.88 g / cm³

Since the density of water is greater than the density of the boat ( 1 > 0.88) then that means, the boat will NOT sink.

B.

4 0
3 years ago
Read 2 more answers
The___ of an element is the number of
castortr0y [4]

Answer:atomic mass, neutrons in the nucleus

Explanation: because I remember from when I took honors chemistry last year we learned about this and it’s called the atomic mass when looking at an atom and the neutrons in the nucleus are effected by it.

8 0
3 years ago
Rearrange the equation for the area of a rectangle (A=l x w) to solve for length, L
Mariana [72]
Not sure if this is what you mean but l= A/w
(length equals area divided by width)
4 0
4 years ago
The total volume required to reach the endpoint of a titration required more than the 50 mL total volume of the buret. An initia
Margaret [11]

Answer:

The endpoint volume is 50.52 ±  0.14 mL

Explanation:

In a titration always is necessary to subtract the blank volume to the titrant volume to obtain the real volume of the titrant. Thus in this case, the total endpoint volume is the sum of the initial volume delivered and the second volume delivered, minus the blank volume:

V = (49.16±0.06 mL) + (1.69±0.04 mL) - (0.33±0.04 mL)

V = (49.16 + 1.69 - 0.33) ± (0.06+0.04+0.04) mL

V = 50.52 ±  0.14 mL

It is necessary to consider the sum of the errors too.

7 0
4 years ago
While doing a lab a student found the density of a piece of pure aluminum to be 2.85 g/cm3 the accepted value for the density of
____ [38]

Answer:

THE PERCENT ERROR IS 5.55 %

Explanation:

To calculate the percent error, we use the formula:

Percent error = Found value - accepted value / accepted value * 100

Found value = 2.85 g/cm3

Accepted value = 2.70 g/cm3

Solving for the percent error, we have:

Percent error = 2.85 g/cm3 - 2.70 g/cm3 / 2.70 g/cm3 * 100

Percent error = 0.15 / 2.70 * 100

Percent error = 0.05555 * 100

Percent error = 5.55 %

In conclusion, the percent error is 5.55 %

3 0
4 years ago
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