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riadik2000 [5.3K]

3 years ago

9

Calculate the density of mercury of it pours exactly 22.5 mL into a graduated cylinder and has a mass of 316 g.

1 answer:

Rasek [7]3 years ago

8 0

Density = mass/volume = 316/22.5 = 14.045g/mL.

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WHAT IS A ROCK?

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Equation balancing

meriva

<u>Answer:</u>

<u>For a:</u> The balanced equation is $2S(s)+3O_2(g)\rightarrow 2SO_3(g)$

<u>For c:</u> The balanced equation is $2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)$

<u>Explanation:</u>

A balanced chemical equation is one where all the individual atoms are equal on both sides of the reaction. It follows the law of conservation of mass.

<u>For (a):</u>

The given unbalanced equation follows:

$S(s)+O_2(g)\rightarrow SO_3(g)$

To balance the equation, we must balance the atoms by adding 2 infront of both $S(s)$ and $SO_3$ and 3 in front of $O_2$

For the balanced chemical equation:

$2S(s)+3O_2(g)\rightarrow 2SO_3(g)$

<u>For (b):</u>

The given balanced equation follows:

$2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)$

The given equation is already balanced.

<u>For (c):</u>

The given unbalanced equation follows:

$2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+H_2O(l)$

To balance the equation, we must balance the atoms by adding 2 infront of $H_2O(l)$

For the balanced chemical equation:

$2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)$

<u>For (d):</u>

The given balanced equation follows:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)$

The given equation is already balanced.

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3 years ago

A sample of an ionic compound NaA, where A- is the anion of a

vitfil [10]

Answer:

a) Kb = 10^-9

b) pH = 3.02

Explanation:

a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:

[NaA] and [A-] = 0.05/0.6 = 0.083 M

Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9

b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:

[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M

pKb = 10^-9

Ka = 10^-5

HA = H+ + A-

Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])

[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0

Clearing [H+]:

[H+] = 0.00095 M

pH = -log([H+]) = -log(0.00095) = 3.02

6 0

3 years ago

The value of the rate constant for a gas phase reaction can be changed by increasing the A. temperature of the reaction vessel.

prohojiy [21]

Answer:

temperature of the reaction vessel

Explanation:

temperature of the reaction vessel

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2 years ago