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3 years ago

Calculate the pOH of a solution if the concentration of hydroxide ions (OH-) is 1.9 x 10-5M?

Chemistry

1 answer:

OLEGan [10]3 years ago

6 0

Answer:

9.28

Explanation:

pOH refers to a measure of hydroxide ions concentration. pOH tells about the alkalinity of a solution. If pOH is less than 7 then aqueous solutions are alkaline, acidic if pOH is greater than 7 and neutral if pOH is equal to 7.

Concentration of the hydroxide ions = 1.9 x 10-5 M

pH = $-log(1.9\times 10^{-5})=4.72$

pOH = 14 - pH

=14 - 4.72 = 9.28

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An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, h

vaieri [72.5K]

Answer:

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

Explanation:

Length of the base = l

Width of the base = w

Height of the pyramid = h

Volume of the pyramid = $V=\frac{1}{3}lwh$

We have:

Rate at which water is filled in cube = $\frac{dV}{dt}= 45 cm^3/s$

Square based pyramid:

l = 6 cm, w = 6 cm, h = 13 cm

Volume of the square based pyramid = V

$V=\frac{1}{3}\times l^2\times h$

$\frac{l}{h}=\frac{6}{13}$

$l=\frac{6h}{13}$

$V=\frac{1}{3}\times (\frac{6h}{13})^2\times h$

$V=\frac{12}{169}h^3$

Differentiating V with respect to dt:

$\frac{dV}{dt}=\frac{d(\frac{12}{169}h^3)}{dt}$

$\frac{dV}{dt}=3\times \frac{12}{169}h^2\times \frac{dh}{dt}$

$45 cm^3/s=3\times \frac{12}{169}h^2\times \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times h^2}$

Putting, h = 4 cm

$\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times (4 cm)^2}$

$=13.20 cm/s$

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

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Please help

ludmilkaskok [199]

Answer: The molecular formula will be $C_3H_9$

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

The empirical formula is $CH_3$

The empirical weight of $CH_3$ = 1(12)+3(1)= 15g.

The molecular weight = 45.0 g/mole

Now we have to calculate the molecular formula:

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{45}{15}=3$

The molecular formula will be= $3\times CH_3=C_3H_9$

Thus molecular formula will be $C_3H_9$

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