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3 years ago

What is the [OH-],[H3O] and PH for the following

Chemistry

1 answer:

malfutka [58]3 years ago

4 0

The values for hydroxide ions, hydronium ions and pH are found in the attached picture.

Explanation:

See the attached picture.

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Which of these reactions is an addition reaction?

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Answer:

https://smartlyhelper.com/chemistry/question14062929

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2 years ago

PLEZ HELP QUICKLY!!!! ILL MARK BRAINLIEST

stellarik [79]

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3 years ago

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Consider the second-order reaction:

kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

Explanation:

$2HI(g)\rightarrow H_2(g)+I_2(g)
$

Rate Law: $k[HI]^2
$

Rate constant of the reaction = k = $6.4\times 10^{-9} L/mol s$

Order of the reaction = 2

Initial rate of reaction = $R=1.6\times 10^{-7} Mol/L s$

Initial concentration of HI = $[A_o]$

$1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2$

$[A_o]=5 mol/L$

Final concentration of HI after t = [A]

t = $4.53\times 10^{10} s$

Integrated rate law for second order kinetics is given by:

$\frac{1}{[A]}=kt+\frac{1}{[A_o]}$

$\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}$

$[A]=0.00345 mol/L$

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

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3 years ago

G.com what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume

tiny-mole [99]

7.379 * 10^(-4) is measured, hence prone to error, either human error or via measuring device. In this case,
100 cm = 1 m is written in stone and is unquestionable.
The density of the gold is 19.3 g/cm^3 and could be an approximation.
The approximation is good to at least one night.

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2 years ago

A mixture consists of 28% oxygen, 14% hydrogen, and 58% nitrogen by volume. A sample of this mixture has a pressure of 4.0 atm i

stepladder [879]

Answer:

C) 1.3 mol

Explanation:

Using gas law we can find the initial moles of the sample of the mixture, as follows:

PV = nRT

PV / RT = n

Where P is pressure: 4.0atm

V is volume: 9.6L

R is gas constant: 0.082atmL/molK

T is absolute temperature: 300K

And n are moles of the gas

PV / RT = n

4.0atm*9.6L / 0.082atmL/molK300K = n

n = 1.56moles of the mixture of the gas are present into the 9.6L container

Now, 14% of this gas is hydrogen that was removed of the system, that is:

1.56mol*14% = 0.22 moles of hydrogen are removed.

Thus, moles of gas that remains in the container are:

1.56mol - 0.22mol = 1.34mol.

Right answer is:

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2 years ago

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