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3 years ago

15

The volume of a gas is 400 mL when the pressure is 1 atm. At the same temperature, what is the pressure at which the volume of t

he gas is 2 L

1 answer:

Tcecarenko [31]3 years ago

4 0

Answer:

the pressure is equal to 0.2atm

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In step 2 of the experimental procedure, you added 6 M sodium hydroxide to the copper(II) nitrate solut form copper(II) hydroxid

kicyunya [14]

Answer:

The solution becomes colorless and a light blue precipitate is formed.

Explanation:

When sodium hydroxide is added to copper(II) nitrate solution, a precipitate of copper(II) hydroxide is formed.

The solution gradually becomes colorless as the light blue precipitate appears.

The reaction is complete (enough sodium hydroxide has been added) when the blue color of the solution completely disappears.

The equation of the reaction is;

Cu(NO3)2(aq) + 2NaOH(aq) → Cu(OH)2(s) + 2NaNO3(aq)

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3 years ago

D. Period 3 element with 6 valence electrons

Kipish [7]

Answer:

Sulfur is an element is Period 3 with 6 valence electrons.

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3 years ago

[06.01]The same amount of heat is removed from 2 kg of water and from 1 kg of water starting at the same temperature. What will

liq [111]

<span>Answer: The 1 kg of water will reach the lowest temperature
</span>
Both of the objects is water so their specific heat should be same. The heat removed from the 2kg of water which is 2 times of mass than the 1kg water. Since the heat removed, both of their temperatures will drops. But 2kg water temperature drops will be half of 1kg water.

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3 years ago

The pH of a 0.78M solution of 4-pyridinecarboxylic acid HC6H4NO2is measured to be 2.53.Calculate the acid dissociation constant

nika2105 [10]

Answer : The value of acid dissociation constant is, $1.1\times 10^{-5}$

Solution : Given,

Concentration pyridinecarboxylic acid = 0.78 M

pH = 2.53

First we have to calculate the hydrogen ion concentration.

$pH=-\log [H^+]$

$2.53=-\log [H^+]$

$[H^+]=2.95\times 106{-3}M$

Now we have to calculate the acid dissociation constant.

The equilibrium reaction for dissociation of (weak acid) is,

$HC_6H_4NO_2\rightleftharpoons C_6H_4NO_2^-+H^+$

initially conc. 0.78 0 0

At eqm. (0.78-x) x x

The expression of acid dissociation constant for acid is:

$k_a=\frac{[C_6H_4NO_2^-][H^+]}{[C_6H_4NO_2]}$

As, $[H^+]=[C_6H_4NO_2^-]=x$

So, $x=2.95\times 106{-3}M$

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

$k_a=\frac{(x)\times (x)}{(0.78-x)}$

$k_a=\frac{(2.95\times 106{-3})\times (2.95\times 106{-3})}{(0.78-2.95\times 106{-3})}$

$K_a=1.1\times 10^{-5}$

Therefore, the value of acid dissociation constant is, $1.1\times 10^{-5}$

7 0

4 years ago

2. Suppose that 21.37 mL of NaOH is needed to titrate 10.00 mL of 0.1450 M H2SO4 solution.

AysviL [449]

Answer:

0.1357 M

Explanation:

(a) The balanced reaction is shown below as:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

(b) Moles of $H_2SO_4$ can be calculated as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $H_2SO_4$ :

Molarity = 0.1450 M

Volume = 10.00 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 10×10⁻³ L

Thus, moles of $H_2SO_4$ :

$Moles=0.1450 \times {10\times 10^{-3}}\ moles$

Moles of $H_2SO_4$ = 0.00145 moles

From the reaction,

1 mole of $H_2SO_4$ react with 2 moles of NaOH

0.00145 mole of $H_2SO_4$ react with 2*0.00145 mole of NaOH

Moles of NaOH = 0.0029 moles

Volume = 21.37 mL = 21.37×10⁻³ L

Molarity = Moles / Volume = 0.0029 / 21.37×10⁻³ M = 0.1357 M

7 0

3 years ago