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3 years ago

Which is the MOST precise measurement this cylinder gives for the volume of liquid it contains?

1 answer:

6 0

Hello! Before I answer to your question, please be sure to include a picture or else a report will be filed for your question would be commenced. You are lucky I have this for my homework tonight and I figured it out. Thank you:

The answer to your question would be as followed:

The most precise measurement for the cylinder you are mentioning is <em><u>B. 43.0mL</u></em>

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A test tube of zinc oxide

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The following charges on individual oil droplets were obtained during an experiment similar to Millikan's. Determine a charge fo

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Answer:

- 1.602 x 10⁻¹⁹coulombs

Explanation:

Charge on individual oil droplet would be multiple of charge on one electron . So we will find out the minimum common factor of given individual charges that is the LCM of all the charges given.

LCM of given charges like 3.204 , 4.806 ,8.01 and 14.42 . We have neglected the power of ten( 10⁻¹⁹) because it is already a common factor to all.

The LCM is 1.602 . So charge on electron is 1.602 x 10⁻¹⁹.

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3 years ago

An atom contains one proton , one electron, and one neutron. Which two particles are most similar in mass

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2 years ago

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A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If

NemiM [27]

Answer:

<h3>25.0 grams is the mass of the steel bar.</h3>

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

$Q_1=-Q_2$

Mass of steel= $m_1$

Specific heat capacity of steel = $c_1=0.452 J/g^oC$

Initial temperature of the steel = $T_1=2.00^oC$

Final temperature of the steel = $T_2=T=21.50^oC$

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2= 105 g$

Specific heat capacity of water= $c_2=4.18 J/g^oC$

Initial temperature of the water = $T_3=22.00^oC$

Final temperature of water = $T_2=T=21.50^oC$

$Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))$

On substituting all values:

$(m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}$

<h3>25.0 grams is the mass of the steel bar.</h3>

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3 years ago

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Answer:

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