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Vsevolod [243]
3 years ago
9

How many total atoms are in 0.830 g of p2o5?

Chemistry
1 answer:
docker41 [41]3 years ago
6 0

Answer:

2.46x10^{22}atoms

Explanation:

Hello,

In this case, we need to compute the atoms of both phosphorous and oxygen, taking into account the following mass-mole-atoms relationship:

Molar,mass=31*2+16*5=142g/mol\\atomsP=0.830gP_2O_5*\frac{1molP_2O_5}{142gP_2O_5} *\frac{2molP}{1molP_2O_5} *\frac{6.022x10^{23}atomsP}{1molP}=7.04x10^{21}atomsP\\atomsO=0.830gP_2O_5*\frac{1molP_2O_5}{142gP_2O_5} *\frac{5molO}{1molP_2O_5} *\frac{6.022x10^{23}atomsO}{1molO}=1.76x10^{22}atomsO

Now, by adding each result, we've got:

atoms=1.76x10^{22}atomsP+7.04x10^{21}atomsO=2.46x10^{22}atoms

Best regards.

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When converting from moles to molecules OR molecules to moles you use what quantitative value?
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Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp
gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200  

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in  j/mol R Ea is

Ea=35.5*1000 j/mol R

ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

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<u><em></em></u>

<u><em>Faraday’s – Second Law of Electrolysis</em></u>

<u><em>Faraday’s second law of electrolysis states that if the same amount of electricity is passed through different electrolytes, the masses of ions deposited at the electrodes are directly proportional to their chemical equivalents.</em></u>

<u><em></em></u>

<u><em>From these laws of electrolysis, we can deduce that the amount of electricity needed for oxidation-reduction depends on the stoichiometry of the electrode reaction.</em></u>

<u><em>The product of an electrolytic reaction depends on the nature of the material being electrolysed and the type of electrodes used. In the case of an inert electrode such as platinum or gold, the electrode does not participate in the chemical reaction and acts only as a source or sink for electrons. While, in the case of a reactive electrode, the electrode participates in the reaction.</em></u>

<u><em></em></u>

<u><em>Hence, different products are obtained for electrolysis in the case of reactive and inert electrodes. Oxidizing and reducing species present in the electrolytic cell and their standard electrode potential too, affect the products of electrolysis.</em></u>

<u><em></em></u>

<u><em>FAQs</em></u>

<u><em>1. What’s a Faraday?</em></u>

<u><em>Ans: The Faraday is an electric charge volume unit without measurements, equal to approximately 6.02 x 10 23 electric charge carriers.</em></u>

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<u><em>2. Why is Faraday’s law important?</em></u>

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<u><em>Ans: Electrolysis is a method of removing iron oxide by passing a small electrical charge through the rusty metal from a battery or battery charger to induce ion exchange while the device is submerged in an electrolyte solution.</em></u>

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<u><em>4. What happens to water during electrolysis?</em></u>

<u><em>Ans: Water’s Electrolysis. By passing an electrical current through it, water can be decomposed. When this happens, an oxidation-reduction reaction is caused by the electrons from the electric current.</em></u>

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<u><em>5. What is the negative electrode called in electrolysis?</em></u>

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<em>Hope it helps!</em>

5 0
2 years ago
Read 2 more answers
Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu
Monica [59]

Answer: The volume of the 1.224 M NaOH solution needed is 26.16 mL

Explanation:

In order to prepare the dilute NaOH solution, solvent is added to a given amount of the NaOH stock solution up to a final volume of 250.0 mL.

Since only solvent is added, the amount of the solute, NaOH, in the dilute solution is the same as in the volume taken from the stock solution.

Molarity (<em>M)</em> is calculated from the following equation:

<em>M</em> = <em>n</em> ÷ <em>V</em>

where <em>n</em> is the number of moles of the solute in the solution, and <em>V</em> is the volume of the solution.

Accordingly, the number of moles of the solute is given by

<em>n</em> = <em>M</em> x <em>V</em>

Now, let's designate the stock NaOH solution and the dilute solution as (1) and (2), respectively . The number of moles of NaOH in each of these solutions is:

<em>n </em>(1) = <em>M </em>(1) x <em>V </em>(1)

<em>n </em>(2) = <em>M </em>(2) x <em>V </em>(2)

As the amount of NaOH in the dilute solution is the same as in the volume taken from the stock solution,

<em>n</em> (1) = <em>n</em> (2)

and

<em>M</em> (1) x <em>V</em> (1)<em> </em>= <em>M</em> (2) x <em>V</em> (2)

For the stock solution, <em>M</em> (1) = 1.244 M, and <em>V</em> (1) is the volume needed. For the dilute solution, <em>M</em> (2) = 0,1281 M, and <em>V</em> (2) = 250.0 mL.

The volume of the stock solution needed, <em>V</em> (1), is calculated as follows:

<em>V</em> (1) = <em>M</em> (2) x <em>V</em> (2) ÷ <em>M</em> (1)

<em>V</em> (1) = 0.1281 M x 250.0 mL ÷ 1.224 M

<em>V </em>(1) = 26.16 mL

The volume of the 1.224 M NaOH solution needed is 26.16 mL.

7 0
3 years ago
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