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2 years ago

what was happening to the H20 when the lines on the graph were mostly flat? What kind of energy was being added to the system?

2 answers:

8 0

Answer: A heating curve graphically represents the phase transitions that a substance undergoes as heat is added to it.

The plateaus on the curve mark the phase changes. The temperature remains constant during these phase transitions.

Water has a high boiling point because of the strong hydrogen bonds between the water molecules; it is both a strong hydrogen bond donor and acceptor.

The first change of phase is melting, during which the temperature stays the same while water melts. The second change of phase is boiling, as the temperature stays the same during the transition to gas

Explanation:

bearhunter [10]2 years ago

6 0

That's latent heat of vaporization.

Water boils at 100°C
The graph increases still 100°C
once it reaches boiling point it yields a straight line because of latent heat after all molecules are evaporated then the graph rises back

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3 years ago

Read 2 more answers

A piece of unknown metal with mass 30 g is heated to 110.0 °C and dropped into 100.0 g of water at 20.0 °C. The final temperatur

Ymorist [56]

Answer: The specific heat of metal is 0.821 J/g°C

Explanation:

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of metal = 30 g

$m_2$ = mass of water = 100 g

$T_{final}$ = final temperature = 25°C

$T_1$ = initial temperature of metal = 110°C

$T_2$ = initial temperature of water = 20.0°C

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$30\times c_1\times (25-110)=-[100\times 4.186\times (25-20)]$

$c_1=0.821J/g^oC$

Hence, the specific heat of metal is 0.821 J/g°C

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3 years ago

What is the majority of 0.50 g of na dissolved in a 1.5 L solution

Oxana [17]

The correct answer is 0.014467 M.

Molarity is defined as the number of moles present in a liter solution, that is, number of moles / liter solution.

The molar mass of sodium (Na) is 23.0 g/mol

Thus, 1 mole of Na contains 23.0 g

Now, x moles of Na contains 0.50 g

Moles = 0.50 × 1 / 23.0

Moles = 0.50 / 23.0

= 0.0217 moles of Na

Molarity = Number of moles / liters of solution

= 0.0217 / 1.5

= 0.014467 M

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