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earnstyle [38]
3 years ago
10

What is the oxidation number of Chlorine in NaClO4?

Chemistry
2 answers:
agasfer [191]3 years ago
4 0

<u>Answer:</u> Oxidation state of chlorine in the given compound is +7.

<u>Explanation:</u>

Oxidation state is defined as the number which is assigned to the element when it gains or loose electrons. If the element gains electron, it will attain a negative oxidation state and if the element looses electrons, it will attain a positive oxidation state.

We take the oxidation state of chlorine atom be 'x'.

Oxidation state of sodium atom = +1

Oxidation state of oxygen atom = -2

Evaluating the oxidation state of chlorine atom:

+1+x+4(-2)=0\\\\x=+7

Hence, the oxidation state of chlorine in the given compound is +7.

Ivahew [28]3 years ago
3 0
The oxidation number of Chlorine in NaClO4 is +7. It can also be estimated by counting the # of bonds to Chlorine.
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__Mg + __Fe2O3 -&gt; __MgO + __Fe
Anika [276]

Answer:

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Explanation:

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7 0
3 years ago
An industrial synthesis of urea obtains 87.5 kg of urea upon reaction of 68.2 kg of ammonia with excess carbon dioxide. Determin
dolphi86 [110]

Answer:

The theoretical yield of urea = <u>120.35kg</u>

The percent yield for the reaction = <u>72.70%</u>

Explanation:

Lets calculate -

The given reaction is -

2NH_3(aq)+CO_2 →CH_4N_2O(aq)+H_2O (l)

Molar mass of urea CH_4N_2O= 60g/mole

Moles of NH_3 = \frac{62.8kg/mole}{17g/mole} (since Moles=\frac{mass  of  substance}{mass of one mole})

                     = 4011.76 moles

Moles of CO_2 = \frac{105kg}{44g/mole}

                = \frac{105000g}{44g/mole}

                = 2386.36 moles

Theoritically , moles of NH_3 required = double the moles of CO_2

    but , 4011.76 , the limiting reagent is NH_3

Theoritical moles of urea obtained = \frac{1 mole CH_4N_2O}{2mole NH_3}\times4011.76 mole NH_3

                                                      = 2005.88mole CH_4N_2O

Mass of 2005.88 mole of CH_4N_2O =2005.88 mole \times\frac{60g CH_4N_2O}{1mole CH_4N_2O}

                                                     = 120352.8g

                                                     120352.8g\times \frac{1kg}{1000g}

                                                     = 120.35kg

Therefore , theroritical yeild of urea = 120.35kg

Now , Percent yeild = \frac{87.5kg}{120.35kg}\times100

                                 72.70%

Thus , the percent yeild for the reaction is 72.70%

8 0
3 years ago
What is the theoretical yield of vanadium, in moles, that can be produced by the reaction of 1.0 mole of V2O5 with 4.0 mole of c
yuradex [85]

Answer:

Theoretical moles of V are 1.6 moles

Explanation:

The theoretical yield of a reaction is defined as the amount of product you would make if all of the limiting reactant was converted into product.

In the reaction:

V2O5(s) + 5Ca(i) → 2V(i) + 5CaO(s)

Based on the reaction, 1 mol of V2O5 needs 5 moles of Ca for a complete reaction. As there are just 4 moles, <em>limiting reactant is Ca.</em> As there are produced 2 moles of V per 5mol of Ca, Theoretical moles of V are:

4 moles of Ca × (2mol V / 5Ca) = <em>1.6 moles of V</em>

<em></em>

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5 0
3 years ago
How did Wegener use fossils as evidence that continents had moved?
Anna35 [415]

Answer:

the fossils on some of the continents were in the same spots as fossils on other continents

Explanation:

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4 0
3 years ago
A sample of Ammonia gas at 650mmHg and 15 degree has a mass of 56.8g.calculate the volume occupied by the gas. Take N=14, H=1,R=
avanturin [10]

Answer:

The first thing we have to do is change and state all the units so that we can use our ideal gas law equation (PV = nRT).

650 mmHg is a pressure unit, we have to convert this to kiloPascals. We know that 760 mmHg gives us 101 kPa.

650 \ mmHg \ * \ \frac{101kPa}{760 mmHg} = 86 \ kPa

P = 86kPa

T = 15°C + 273K = 288K

R (Gas constant) = 8.31 kj/mol  

Molar mass of Ammonia (NH_{3}) = (1 x 3) + (14) = 17g/mol

n (moles) = \frac{mass}{molar \ mass} = \frac{56.8}{17} = 3.34 mol

V = ?

Rearrange the equation to solve for Volume:

V = \frac{nRT}{P}

Substitute the values inside:

V = \frac{(3.34)(8.31)(288)}{(86)} = 93 L (rounded)

<u>Therefore 93 L of volume is occupied by the ammonia gas.</u>

<u></u>

4 0
2 years ago
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