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melisa1 [442]
3 years ago
12

Setting up a one-step unit conversion A student sets up the following equation to convert a measurement. (The ? stands for a num

ber the student is going to calculate.) Fill in the missing part of this equation. 0.080 cm = ?m Check Student Employment Service
Chemistry
1 answer:
Aneli [31]3 years ago
7 0

Answer:

0.0008 m

Explanation:

We are given that 0.080 cm

We have to convert 0.080 cm into meter

To find the value of 0.080 cm in meter we are using unitary method

We know that

100 cm =1 m

1 cm =\frac{1}{100} m

Therefore, 0.080 cm =\frac{1}{100}\times 0.080 m

0.080 cm =\frac{1}{100}\times \frac{80}{1000}m

0.080 cm =\frac{8}{10000}m

0.08cm=0.0008 m

Hence, the value of 0.080 cm is equal to 0.0008 meter .

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A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C・g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('
polet [3.4K]

Answer:

a=5.65cm

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

Q_{Pt}=-Q_{Deu}

We can represent the heats in terms of mass, heat capacities and temperatures:

m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})

Thus, we solve for the mass of platinum:

m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g

Next, by using the density of platinum we compute the volume:

V_{Pt}=\frac{3860.4g}{21.45g/cm^3}\\ \\V_{Pt}=180cm^3

Which computed in terms of the edge length is:

V=a^3

Therefore, the edge length turns out:

a=\sqrt[3]{180cm^3}\\ \\a=5.65cm

Best regards.

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3 years ago
Before measuring the absorbance of a solution with the ocean optics spectrophotometer, it is important to run a blank sample of
Strike441 [17]
I think the most appropriate answer is: the solvent being used in the experiment
<span>To correct for any light absorption not originating from the solute you will need to calibrate the tools with a solution that most similar to the sample.
Blank covete or standard solution can be used, but it was not ideal. By using the solvent as calibration, you can remove the reading from the solvent so your result only comes from the sample.

</span>
5 0
3 years ago
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