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3 years ago

Which portion of a molecule of F2O has partial positive charge?

Chemistry

1 answer:

sweet-ann [11.9K]3 years ago

5 0

Answer:

The difference in electronegativity between fluorine (4.0) and hydrogen (2.1) is quite high, so the shared electrons spend much more time in the vicinity of the fluorine atom. As a result, fluorine carries a partial negative charge in this molecule, whereas hydrogen carries a partial positive charge

Explanation:

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Solubility refers to the ability of a substance to dissolve in a solution. please select the best answer from the choices provid

natima [27]

Solubility is the property of the solute to dissolve itself in a solvent. The states of the solute and solvent may range from solid, liquid, and gas. Since the statement above states a solution instead of a solvent. The answer is false. solution is composed both of the dissoved solute and solvent.

6 0

3 years ago

If 5.48 g of CuNO3 is dissolved in water to make a 0.840 M solution, what is the volume of the solution in milliliters?

Ainat [17]

Answer:

34.8 mL

Explanation:

First we convert 5.48 g of CuNO₃ into moles, using its molar mass:

5.48 g CuNO₃ ÷ 187.56 g/mol = 0.0292 mol

Now we can calculate the volume of the solution, using the definition of molarity:

Molarity = moles / liters

0.840 M = 0.0292 mol / liters

liters = 0.0348 L

Finally we convert L into mL:

0.0348 L * 1000 = 34.8 mL

5 0

3 years ago

How many significant figures are in the measurement 603.040 g?

hichkok12 [17]

6.
hope this helps!!!

3 0

3 years ago

In some regions of the southwest United States, the water is very hard. For example, in Las Cruces, New Mexico, the tap water co

myrzilka [38]

Answer:

a) $46.734L$ of hard water is require to produce $45L$ of pure water.

b) Same water would not work for purifying sea water as it would require so much of energy.

Explanation:

Osmotic pressure of a solution is calculated by the formula,

Π= $iMRT$

Where,

Π is the osmotic pressure

$i$ is van't hoff factor

$M$ is molar concentration (mol/L)

$R$ is gas constant

$T$ is Temperature in Kelvin

Now,

$MgCO_{3}\rightarrow Mg^{2+}+ CO_{3}^{2-}$

So, according to the above equation $i=2$

Calculation concentration at which reverse osmosis will stop

$M=\frac{ Π }{iRT}$

$M=\frac{8atm}{2\times0.082L atm mol^{-1}K^{-1} \times 300K} \\M= 0.162M$

Calculating Initial concentration that is moles of $MgCO_{3}$

$n=\frac{given mass}{molar mass} \\\\n=\frac{560 \times 10^{-6}g}{84.32g/mol}\\\\N=6.642\times10^{-6}moles$

Calculating molarity

$M= 6.642 \times 10^{-6}\times 10{^3}\\M=6.642\times10^{-3}M$

Total number of moles in total volume must remain the same,

$C_{1}V_{1}=C_{2}V_{2}$

$0.006642M \times V_{1}=0.162M \times V_{2}$

$\frac{V_{1}}{V_{2}}= \frac{0.162}{0.006.642}=25.953$

$V_{1}=25.953 V_{2}$

Also, $V_{1}=45L$

$V_{2}=\frac{45}{25.953}\\ V_{2}= 1.734L$

So, net Volume,

$V=45L+1.734L=46.734L$

So, $46.734L$ is require to produce $45L$ of pure water.

Using the equation for $0.60M NaCl$ as well,

$C_{1}V_{1}=C_{2}V_{2}$

$C_{1}=0.60M\\V_{1}=45L\\C_{2}=0.162M\\V_{2}=?$

$V_{2}=\frac{0.60 \times 45}{0.167}\\V_{2}= 166.66L$

$Net water=166.66+45L=211.66L$

So, the same water would not work for purifying sea water as it would require so much of energy, as we require 211.66L of water to produce 45L of water.

7 0

3 years ago

Would you expect a butane lighter to work in winter when the temperature outdoors is 25 ∘F?

kogti [31]

No, I would not expect the lighter to work at 25 °F.

25 °F ≈ -4 °C

The boiling point of butane is -1 °C.

At -4 °C, the vapour pressure of butane is not great enough to push back against atmospheric pressure.

The lighter won't work.

7 0

3 years ago