All categories

3 years ago

How many significant figures are in 0.08260 L? 4 5 6

Chemistry

2 answers:

mars1129 [50]3 years ago

6 0

Answer:

Explanation:

Count the digits after the first non zero number, any number after that (even zero) counts.

Hope that helped!!! k

Anni [7]3 years ago

5 0

Answer:

$\boxed{4}$

Explanation:

$\sf Leading \ zeros \ are \ not \ significant.$

$\sf Trailing \ zeros \ are \ significant.$

$\sf Non\ zeros \ are \ significant.$

$0.08260 \sf \ has \ 4 \ significant \ figures.$

You might be interested in

How many grams of water at 50◦c must be added to 16 grams of ice at −12◦c to result in only liquid water at 0◦c?

melisa1 [442]

When water at 50 C is added to ice at -12 C, heat is transferred from hot water to ice.

- Heat given out by water = Heat absorbed by ice

Calculating the heat released by hot water:

$q = m Cwater$ ΔT $q = m (4.184\frac{J}{(g.^{0}C)} )$ $(0^{0}C-50^{0}C)$

Calculating heat absorbed by 16 g of ice: Ice at $-12^{0}C$ is converted to ice at $O^{0}C$ and then ice at $O^{0} C$ to water at $0^{0}C$

$q = m Cice$ ΔT + $m (Heat of fusion)$

$q = 16 g(2.11\frac{J}{g.^{0}C})(0^{0}C - (-12^{0}C))$ + $16 g (333.55\frac{J}{g})$

q = 405.12 J +5336.8 J =5741.92 J

- Heat given out by water = Heat absorbed by ice

-( $m(4.184\frac{J}{g.^{0}C})(0^{0}C-50^{0}C) = 5741.92 J$

m = 27.4 g

Therefore, 27.4 g water at $50^{0}C$ must be added to 16 g of ice at $-12^{0}C$ to convert to liquid water at $0^{0}C$

4 0

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:

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

For process 1:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_s=0.97J/g.K\\T_2=-144^oC\\T_1=-155^oC\\\Delta T=[T_2-T_1]=[-144-(-155)]^oC=11^oC=11K$

Putting values in equation 1, we get:

$q_1=42.0g\times 0.97J/g.K\times 11K\\\\q_1=448.14J$

For process 2:

We are given:

$m=42.0g\\\Delta H_{fusion}=5.02kJ/mol=\frac{5.02kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=109.13J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 109.13J/g\\\\q_2=4583.5J$

For process 3:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=-144^oC\\\Delta T=[T_2-T_1]=[78-(-144)]^oC=222^oC=222K$

Putting values in equation 1, we get:

$q_3=42.0g\times 2.3J/g.K\times 222K\\\\q_3=21445.2J$

For process 4:

We are given:

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

Putting values in equation 2, we get:

$q_4=42.0g\times 773.04J/g\\\\q_4=32467.7J$

Total heat required = $[q_1+q_2+q_3+q_4]$

Total heat required = $[448.14+4583.5+21445.2+32467.7]J=58944.5J$

Hence, the total heat required is 58944.5 J

8 0

3 years ago

Calculate the poh of this solution. round to the nearest hundredth. ph = 1.90 poh =

Viktor [21]

Answer:

The answer is 5.10

Explanation:

<h3>Given;</h3>

pH = 1.9

pOH = ?

We know that

pH + pOH = 7

pOH = 7 – pH

pOH = 7 – 1.90

pOH = 5.10

Thus, The pOH of the solution is 5.10

6 0

1 year ago

A 2.0-liter aqueous solution contains a total of 3.0 moles of dissolved NH4Cl at 25°C and standard pressure.

vladimir1956 [14]

The molarity is a concentration unit which defined as the number of moles of solute divided by the number of liters of solution. So the molarity of the solution is 3/2=1.5 mol/L.

7 0

2 years ago

Read 2 more answers

What is love and why are people scared to show love even if they like boys or girls

myrzilka [38]

Because society has shaped our mind on what we shall think about things

6 0

3 years ago

Read 2 more answers