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3 years ago

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What would be the formula for diphosphorus pentoxide? A. PO5 B. P2O5 C. P5O D. P5O2

2 answers:

Dovator [93]3 years ago

6 0

I am sure, the answer is variant B.

Nookie1986 [14]3 years ago

4 0

heeeeeeyyyyyy yalllllll the answer is B!!!!!!!!!!!

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In ionic bonding, atoms__.

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In ionic bonding, atoms SHARE ELECTRONS

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3 years ago

000333 g fluorescein (332.32 g/mol) is dissolved in 225 ml solution of ethanol. the density of ethanol is 0.785 g/ml. what is th

andrey2020 [161]

1) Molarity

M = n / V
n: number of moles of solute
V: volume of the solution in liters

n = mass / molar mass = 0.000333 g / 332.32 g / mol = 1*10 ^ - 6 moles

V = 225 ml * 1 liter / 1000 ml = 0.225 liter

M = 10^-6 mol / 0.225 liter = 0.00000444 M

2) ppm

ppm = parts per million

grams of solute: 0.000333 g

grams of solution = volume * density = 225 ml * 0.785 g / ml = 176.625 g

ppm = [0.00033 g / 176.625 g] * 1,000,000 = 1.868 ppm

8 0

3 years ago

Read 2 more answers

Most thermal conductor are? A.plastic B.glass C.rubber D. Metal

Alex17521 [72]

It is metal that is why you cook things mainly with metal pots

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3 years ago

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Calculate the pH of the solution formed when 45.0 mL of 0.100M NaOH solution is added to 50.0 mL of 0.100M CH3COOH (Ka for aceti

pav-90 [236]

Answer:

pH of soltion will be 5.69

Explanation:

The pH of the solution will be due to excessive acid left and the salt formed. Thus, it will form a buffer solution.

The pH of buffer solution is calculated from Henderson Hassalbalch's equation, which is:

$pH=pKa+log(\frac{[salt]}{[acid]} )$

$pKa=-log(Ka)$

$pKa=-log(1.8X10^{-5})=4.74$

The moles of acid taken :

$moles=molarityXvolume=0.1X50=5mmol$

The moles of base taken:

$moles=molarityXvolume=0.1X45=4.5mmol$

The moles of acid left after reaction :

$5-4.5=05mmol$

The moles of salt formed = 4.5mmol

Putting values in equation

$pH=pKa+log(\frac{[salt]}{[acid]} )=4.74+log(\frac{4.5}{0.5})=5.69$

8 0

3 years ago

Lead(II) nitrate is added slowly to a solution that is 0.0800 M in CT ions. Calculate the concentration of Pb2+ ions (in mol/L)

charle [14.2K]

Answer : The concentration of $Pb^{2+}$ ion is 0.0375 M.

Explanation :

The balanced equilibrium reaction will be:

$Pb^{2+}+2Cl^-\rightleftharpoons PbCl_2$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Pb^{2+}][Cl^-]^2$

Now put all the given values in this expression, we get:

$2.40\times 10^{-4}=[Pb^{2+}]\times (0.0800)^2$

$[Pb^{2+}]=0.0375M$

Therefore, the concentration of $Pb^{2+}$ ion is 0.0375 M.

8 0

3 years ago