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3 years ago

Boiling Point Electrical conductivity in aqueous solution Solubility in H2O (grams of solid in 100 grams H2O)

Chemistry

2 answers:

Allisa [31]3 years ago

8 0

Answer:

A) A contains sugar and B contains salt, because salt is ionic and a good conductor of electricity.

Explanation:

Because sugar is not a good conductor of electricity

sleet_krkn [62]3 years ago

6 0

Answer:

A) A contains sugar and B contains salt, because salt is ionic and a good conductor of electricity.

Explanation:

A contains sugar and B contains salt, because salt is ionic and a good conductor of electricity.. Covalent compounds do not form charged ions in solution, thus they are poor conductors of electricity, while ionic compounds do produce ions in solution and are good conductors. Sugar is a covalent compound, thus it would be a poor conductor in solution.

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4 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 = $T_2=T=21.50^oC$

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

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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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4 years ago

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