All categories

3 years ago

12. Which of the planets in the solar system

1 answer:

4 0

Between 1858 and 1920, Mars was thought by many, including some scientists, to be very similar to Earth, only drier with a thick atmosphere, similar axial tilt, orbit and seasons as well as a Martian civilization that had built great Martian canals.

You might be interested in

If I titrate an acid with 255mL of 3.4 M NaOH (base) and reach the equivalence point, how many moles of H+ ions were in the acid

FromTheMoon [43]

Answer:

Molar Concentration = $\frac{Moles of Solute}{Litres of Solution}$

= $\frac{3.4}{0.255}$

= 13.33

No. of H+ ions present = 13.33

pH value = - log[13.33]

= -1.12

Explanation:

The equivalence point, or stoichiometric point, of a substance response is the point at which synthetically identical amounts of reactants have been blended. As such, the moles of corrosive are equal to the moles of base, as per the condition (this doesn't really infer a 1:1 molar proportion of acid:base, simply that the proportion is equivalent to in the condition). It tends to be found by methods for a marker, for instance phenolphthalein or methyl orange. The endpoint (identified with, however not equivalent to the equivalence point) alludes to the point at which the marker changes shading in a colorimetric titration.

7 0

3 years ago

A 0.100 M oxalic acid, HO2CCO2H, solution is titrated with 0.100 M KOH. Calculate the pH when 25.00 mL of oxalic acid solution i

____ [38]

Answer:

pH = 4.09

Explanation:

molarity of oxalic acid in the solution

= 0.1 x 25 / (25 + 35)

= 0.0417 M

molarity of NaOH in the solution

= 0.1 x 35 / (25 +35)

= 0.0583 M

H2C2O4 + NaOH -------------------> NaHC2O4 + H2O

0.0417 0.0583 0 0

0 0.0166 0.0417

now second acid -base titration

NaHC2O4 + NaOH -------------------> Na2C2O4 + H2O

0.0417 0.0166 0 0

0.0251 0 0.0166 ---

now

pH = pKa2 + log [Na2C2O4 / NaHC2O4]

pH = 4.27 + log (0.0166 / 0.0251)

pH = 4.09

3 0

3 years ago

In caves, dripping groundwater containing calcium bicarbonate can form stalactites and stalagmites. When the water is exposed to

Harrizon [31]

The right answer for the question that is being asked and shown above is that: "E. This process involves breaking and reshaping of substances, which is an an important clue to chemical change." The process of a chemical change is that it involves breaking and reshaping of substances, which is an an important clue to chemical change.

6 0

3 years ago

Read 2 more answers

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:

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