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3 years ago

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6th grade science help me please (:

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S_A_V [24]3 years ago

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A. The atmosphere is layered

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Benzoic acid, c6h5cooh, has a ka = 6.4 x 10-5. what is the concentration of h3o+ in a 0.5 m solution of benzoic acid? 3.2 x 10-5

FinnZ [79.3K]

The answer for this issue is:
The chemical equation is: HBz + H2O <- - > H3O+ + Bz-
Ka = 6.4X10^-5 = [H3O+][Bz-]/[HBz]
Let x = [H3O+] = [Bz-], and [HBz] = 0.5 - x.
Accept that x is little contrasted with 0.5 M. At that point,
Ka = 6.4X10^-5 = x^2/0.5
x = [H3O+] = 5.6X10^-3 M
pH = 2.25
(x is without a doubt little contrasted with 0.5, so the presumption above was OK to make)

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3 years ago

Weight and mass are essentially the same measurements.<br> True or false?

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The answer to this question is False

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What is the factor of 1024

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QUESTION 3<br><br> A large system of stars held in place by mutual gravitation. IT

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3 years ago

The benzoate ion, c6h5coo− is a weak base with kb=1.6×10−10. how many moles of sodium benzoate are present in 0.50 l of a soluti

lyudmila [28]

$NaC6H5COO \rightarrow Na{^{+}} + C6H5COO^{-}$

Here the base is a benzoate ion, which is a weak base and reacts with water.

$C6H5COO^{-}(aq) + H2O (l)\leftrightarrow C6H5COOH(aq)+ OH^{-}(aq)$

The equation indicates that for every mole of OH- that is produced , there is one mole of C6H5COOH produced.

Therefore [OH-] = [C6H5COOH]

In the question value of PH is given and by using pH we can calculate pOH and then using pOH we can calculate [OH-]

pOH = 14 - pH

pH given = 9.04

pOH = 14-9.04 = 4.96

pOH = -log[OH-] or $[OH^{-}] = 10^{^{-pOH}}$

$[OH^{-}] = 10^{^{-4.96}}$

$[OH^{-}] = 1.1\times 10^{-5}$

The base dissociation equation kb = $\frac{Product}{Reactant}$

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

H2O(l) is not included in the 'kb' equation because 'solid' and 'liquid' are taken as unity that is 1.

Value of Kb is given = $1.6\times 10^{-10}$

And value of [OH-] we have calculated as $1.1\times 10^{-5}$ and value of C6H5COOH is equal to OH-

Now putting the values in the 'kb' equation we can find the concentration of C6H5COO-

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

$1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]}$

$[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}$

$[C6H5COO^{-}] = 0.76 M$ or $0.76\frac{mol}{L}$

So, Concentration of NaC6H5COO would also be 0.76 M and volume is given to us 0.50 L , now moles can we calculated as : Moles = M X L

Moles of NaC6H5COO would be = $0.76(\frac{mol}{L}) \times (0.50L)$

Moles of NaC6H5COO (sodium benzoate) = 0.38 mol

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3 years ago

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