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3 years ago

Why is boiling water a physical change

Chemistry

1 answer:

Harrizon [31]3 years ago

6 0

Answer:

Boiling water is an example of a physical change and not a chemical change because the water vapor still has the same molecular structure as liquid water (H2O).

Explanation:

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A sample of sand, 5cm in diameter and 10 cm long, was prepared at a porosity of 50% in a constant head apparatus. The total head

GarryVolchara [31]

Answer:

$K = 2.037*10^{-3} m/s$

$V_s = 0.0122 \ m/s$

Explanation:

Given that;

diameter (d) = 10cm/2 = 0.1m/2 = 0.05 m

length (l) = 10 cm = 0.1 m

porosity = 50%

height (h) = 30 cm = 0.3 m

time (t) = 5 s

volume (v) = 60 cm³ = 60 × 10⁻⁶ m³

Q (flow rate) = $\frac{v}{t}$

Q = $\frac{60*10^{-6} m^3}{5}$

Q = $12*10^{-6} m^3 /sec$

From constant head method, we use the relation K = $(\frac{Q*L}{A*h})$ to determine the hydraulic conductivity ; we have:

$K = \frac{12*10^{-6}*0.1}{\frac{\pi}{4}0.05^2*0.3}$

$K = 0.002037\\\\K = 2.037*10^{-3} m/s$

Seepage velocity $V_s = \frac{velocity }{porosity}$

where; velocity = $K*i$

= $(2.037*10^{-3}*)(\frac{0.3}{0.1})$

= $6.111*10^{-3} m/s$

$V_s = \frac{6.111*10^{-3}}{0.5}$

$V_s = 0.0122 \ m/s$

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3 years ago

Which alkaline earth metal is part of the reaction process of photosynthesis answers?

Brrunno [24]

Magnesium????
i hope this helps

5 0

3 years ago

the element weighed 3.555 grams when it was burned in oxygen an oxide compound was produced. the product mass was 4.666 grams. W

vodka [1.7K]

Idk really but i hope you figure out the answer im not on that yet.

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3 years ago

Explain how you determine molar mass of Ca(No3)2

igor_vitrenko [27]

Answer:

164

1st step we will write desperate molar mass of each element

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3 years ago

Ethanol (C2H5OH) melts a - 144 oC and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vapo

hammer [34]

Answer:

For a: The total heat required is 36621.5 J

For b: The total heat required is 58944.5 J

Explanation:

For a:

To calculate the heat required at different temperature, we use the equation:

$q=mc\Delta T$ .........(1)

where,

q = heat absorbed

m = mass of substance

$c$ = specific heat capacity of substance

$\Delta T$ = change in temperature

To calculate the amount of heat required at same temperature, we use the equation:

$q=m\times \Delta H$ ........(2)

where,

q = heat absorbed

m = mass of substance

$\Delta H$ = enthalpy of the reaction

The processes involved in the given problem are:

$1.)C_2H_5OH(l)(35^oC)\rightarrow C_2H_5OH(l)(78^oC)\\2.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

For process 1:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=35^oC\\\Delta T=[T_2-T_1]=[78-35]^oC=43^oC=43K$

Putting values in equation 1, we get:

$q_1=42.0g\times 2.3J/g.K\times 43K\\\\q_1=4153.8J$

For process 2:

We are given:

Conversion factor: 1 kJ = 1000 J

Molar mass of ethanol = 46 g/mol

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{35.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 773.04J/g\\\\q_2=32467.7J$

Total heat required = $[q_1+q_2]$

Total heat required = $[4153.8J+32467.7J]=36621.5J$

Hence, the total heat required is 36621.5 J

For b:

The processes involved in the given problem are:

$1.)C_2H_5OH(s)(-155^oC)\rightarrow C_2H_5OH(s)(-144^oC)\\2.)C_2H_5OH(s)(-144^oC)\rightarrow C_2H_5OH(l)(-144^oC)\\3.)C_2H_5OH(l)(-144^oC)\rightarrow C_2H_5OH(l)(78^oC)\\4.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$