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3 years ago

The molality is temp independent whereas molarity is temperature dependent. Justify.

1 answer:

6 0

Molality is obtained by dividing the number of moles of solute by the mass in kilogram of the solvent. None of the dimensions is dependent in temperature. On the other hand, molarity is obtained by dividing the number of moles of solute by the volume in liters of the solution. Volume is temperature dependent.

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Bruce left some butter out in his kitchen. Bright sunlight came in through the window and melted the butter. Heat energy from th

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The correct answer is D radiation here is why solar radiation simply refer to energy from the sun.

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What is always formed when a strong acid solution reacts with a strong base solution?

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Answer:

water and salt

Explanation:

A neutralization reaction is when an acid and a base react to form water and a salt and involves the combination of H+ ions and OH- ions to generate water. The neutralization of a strong acid and strong base has a pH equal to 7.

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2 years ago

Question 9 Multiple Choice Worth 4 points)

Rufina [12.5K]

Answer:

Explanation:

Would you base your career options using science?

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2 years ago

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An object has a density of 16.3 g/mL and a volume of 0.46 L. Calculate the

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Answer:

16.53 pounds

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2 years ago

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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