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Sonja [21]
3 years ago
15

10) A student’s calculation w as found to have a 15.6% error, and the actual value w as determined to be 25.7 mL. W hat are the

tw o possible values for the student’s experimental measurement?
Chemistry
1 answer:
goblinko [34]3 years ago
7 0

Answer:

The actual value = 25.7 ml

The error = 15.6%

This means that the value of error = 0.156 * 25.7 = 4.0092 ml

Now, the error percentage found means that the student got the value either greater than the actual one with the value of error or less than the actual one with the value of error.

This means that the two possible readings are:

either : 25.7 + 4.0092  = 29.7092 ml

or : 25.7 - 4.0092  = 21.6908 ml

Explanation:

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Using henry's law, calculate the molar concentration of o2 in the surface water of a mountain lake saturated with air at 20 ∘c a
s2008m [1.1K]

Answer: The molar concentration of oxygen gas in water is 1.43\times 10^{-7} mol/L.

Explanation:

Partial pressure of the O_2gas = 685 torr = 0.8905 bar

1 torr = 0.0013 bar

According Henry's law:

p_{o_2}=K_H\times\chi_{O_2}

Value of Henry's constant of oxygen gas at 20 °C in water = 34860 bar

0.0013=34860 bar\times \chi_{O_2}

\chi_{O_2}=\frac{0.8905 bar}{34680 bar}=2.56\times 10^{-5}

Let the number of moles of O_2 gas in 1 liter water be n.

1 Liter water = 1000 g of water

Moles of water in 1 L n_w=\frac{1000 g}{18 g/mol}=55.55 mol

\chi_{O_2}=\frac{n}{n+n_w}

2.56\times 10^{-5}=\frac{n}{n+55.55}

n=1.43\times 10^{-7} moles

Molarity=\frac{\text{Moles of}O_2}{Volume}

Molar concentration of oxygen gas in 1 L of water:

=\frac{1.43\times 10^{-7} moles}{1 L}=1.43\times 10^{-7} mol/L

The molar concentration of oxygen gas in water is 1.43\times 10^{-7} mol/L.

4 0
3 years ago
A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt
Nadusha1986 [10]

Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt

integrating factor

=e^{\int\limits (1/40) \, dt}

integrating factor =e^{(1/40)t}

multiply on both sides by  =e^{(1/40)t}

dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t

integrate on both sides

\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})

b)

after long period of time means t - > ∞

{t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
6 0
3 years ago
Calculate the molar mass of the acid.
JulijaS [17]
Citric acid has the molecular formula C6H8O7 so you can add the molar masses of the elements from the periodic table. C has a molar mass of 12.01 g/mol, H has 1.01 g/mol and O has 15.999 g/mol. Now you calculate the total molar mass= (6*12.01 + 8*1.01 + 7*15.999). This yields a molar weight of 192.124 g/mol (anhydrous)
3 0
3 years ago
The essential oil found in cloves, eugenol, can be isolated by steam distillation because it is insoluble in water and has a mea
Damm [24]
We know that:
Molar Mass H2O: 18 g/mol 
<span>Molar Mass of Eugenol: 164 g/mol </span>
<span>Boiling point of H2O: 100 degrees C </span>
<span>Boiling point of Eugenol: 254 degrees C </span>
<span>Density of water: 1.0 g/mL </span>
<span>Density of Eugenol: 1.05 g/mL </span>

<span>Using formula:
V= [mole fraction x molar mass] / density </span>

<span>mH20: 0.9947 * 18
          = 17.9046 / 1 g/mL
          = 17.9046 </span>
<span>morg: 0.0053 * 164  
        = 0.8692/ 1.05 g/mL
        = 0.8278 </span>

<span>V% = Vorg/(Vorg + VH2O) * 100 </span>
<span>(0.8278/18.7324) * 100 = 4.419% </span>

Yotal volume = 30 mL; therefore, 
<span>0.0442 = (volume eugenol/30) </span>

<span>(m eug/mH2O) = (peug*164/pH2O*18) </span>
<span>(m eug/30) = (4*164/760*18) </span>
<span>m eug = about 1.44g and </span>
<span>
volume = mass/density
            = 1.44/1.05
            = about 1.37 mL </span>
6 0
2 years ago
How many structures are possible for a square planar molecule with a formula of ax3y?
Galina-37 [17]

Answer:

Explanation:

Depending upon the relative arrangements of XandY X a n d Y , the square planar molecule AX3Y A X 3 Y shows only the following structure: Hence, only one structure is possible for a square planar molecule with a formula of AX3Y A X 3 Y .

5 0
2 years ago
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