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mrs_skeptik [129]

2 years ago

11

Balance the following reaction. A coefficient of "1" is understood. Choose option "blank" for the correct answer if the coeffici

ent is "1".
NH3 + F2 → N2F4 + HF

2 answers:

Sveta_85 [38]2 years ago

8 0

2 NH3 + 5F2 = blank N2F4 + 6 HF

erica [24]2 years ago

3 0

Answer : The balanced chemical reaction will be,

$2NH_3+5F_2\rightarrow N_2F_4+6HF$

Explanation :

Balanced chemical reaction : It is defined as the chemical reaction in which the number of individual atoms of an element in reactant side always be equal to the number of individual atoms of an element in product side.

The given balanced reaction will be,

$2NH_3+5F_2\rightarrow N_2F_4+6HF$

By the stoichiometry we can say that, 2 moles of $NH_3$ react with 5 moles of $F_2$ to give 1 mole of $N_2F_4$ and 6 moles of $HF$ as a product.

Hence, the balanced chemical reaction will be,

$2NH_3+5F_2\rightarrow N_2F_4+6HF$

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In the adjoining figure, the circles with the + sign represent protons and the empty circles represent neutrons. Determine which

Natali [406]

Answer: option C. 2 and 3.

Explanation:

1) Isotopes are atoms of a same element with different number of neutrons. That means that the isotopes of a same element have the same number of protons, since the number of protons is what identify an element.

2) For example, all the atoms of oxygen have 8 protons. But isotope oxygen-16 has 8 neutrons, while oxygen-18 has 10 neutrons.

3) In the figure there are 3 different atoms:

i) atom # 1 has 5 protons and 7 neutrons

ii) atom # 2 has 6 protons and 7 neutrons

iii) atom # 3 has 6 protons and 8 neutrons.

4) Hence the atoms with the same number of protons and different number of neutrons are the #2 and the # 3. So, they are the isotopes of the same element.

4 0

2 years ago

Read 2 more answers

The chlorination of methane occurs in a number of steps that results in the formation of chloromethane and hydrogen chloride. Th

kenny6666 [7]

Answer:

Total pressure = 0,806 atm

Partial pressure of CH₄: 0,037 atm

Partial pressure of Cl₂: 0,396 atm

Partial pressure of CH₃Cl: 0,125 atm

Partial pressure of HCl: 0,125 atm

Partial pressure of Cl⁻: 0,125 atm

Explanation:

For the reaction:

2CH₄(g)+3Cl₂(g)⟶2CH₃Cl(g)+2HCl(g)+2Cl⁻(g)

295 mL≡ 0,295L of methane at STP are:

n = PV/RT

P = 1 atm; V = 0,295L; R = 0,082atmL/molK; T = 273K.

moles of methane: 0,0132 moles

For 725 mL of chlorine ≡ 0,725L

moles of chlorine at STP are: ≡ 0,0324 moles

For a complete reaction of 0,0132 moles of CH₄:

0,0132 mol CH₄× $\frac{3molCl_{2}}{2 molCH_{4}}$ = <em>0,0198 moles</em>

The reaction reaches 77%, moles of Cl₂ that react are: 0,0198×77% = 0,0153 mol

As you have 0,0324 moles of Cl₂, moles that will not react are:

0,0324 - 0,0153 = <em>0,0171 mol Cl₂</em>

As the reaction reaches 77% completion, moles of CH₄ that react are:

0,0132×77% =<em> 0,0102 moles of CH₄ And the moles that don't react are </em><em>0,00300 mol</em>

Thus, moles of each compound are:

0,0102 moles of CH₄× $\frac{3Cl_{2}}{2 molCH_{4}}$ = <em>0,0153 mol + 0,0171 mol = 0,0324 mol Cl₂</em>

0,0102 moles of CH₄× $\frac{2CH_{3}Cl}{2 molCH_{4}}$ = <em>0,0102 mol CH₄</em>

0,0102 moles of CH₄× $\frac{2HCl}{2 molCH_{4}}$ = <em>0,0102 mol HCl</em>

0,0102 moles of CH₄× $\frac{2Cl^{-}}{2 molCH_{4}}$ = <em>0,0102 mol Cl⁻</em>

Total pressure using:

P = nRT/V

Where: n = 0,0102mol×3+0,0324mol + 0,0030mol = 0,0660mol; R = 0,082 atmL/molK; T = 298K; V = 2L

Total pressure = 0,806 atm

Partial pressure of CH₄: 0,037 atm

Partial pressure of Cl₂: 0,396 atm

Partial pressure of CH₃Cl: 0,125 atm

Partial pressure of HCl: 0,125 atm

Partial pressure of Cl⁻: 0,125 atm

<em>-To obtain partial pressure you change the moles for each compound-</em>

<em />

I hope it helps!

5 0

2 years ago

How many grams of NH3 can be prepared from the synthesis of 77.3 grams of nitrogen and 14.2 grams of hydrogen gas?

lbvjy [14]

Answer:

80.41 g

Explanation:

Data Given:

Mass of Nitrogen (N₂) = 77.3 g

Mass of Hydrogen (H₂) = 14.2 g

many grams of NH₃ = ?

Solution:

First we look at the balanced synthesis reaction

N₂ + 3 H₂ ------—> 2 NH₃

1 mol 3 mol

As 1 mole of Nitrogen react with 3 mole of hydrogen

Convert moles to mass

molar mass of N₂ = 2(14) = 28 g/mol

molar mass of H₂ = 2(1) + 2 g/mol

Now

N₂ + 3 H₂ ------—> 2 NH₃

1 mol (28 g/mol) 3 mol(2g/mol)

28 g 6 g

28 grams of N₂ react with 6 g of H₂

So

if 28 grams of N₂ produces 6 g of H₂ so how many grams of N₂ will react with 14.2 g of H₂.

Apply Unity Formula

28 g of N₂ ≅ 6 g of H₂

X g of N₂ ≅ 14.2 g of H₂

Do cross multiply

X g of N₂ = 28 g x 14.2 g / 6 g

X g of N₂ = 66.3 g

As we have given with 77. 3 g of N₂ but from this calculation we come to know that 66.3 g will react with 14.2 g of hydrogen and the remaining 10 g N₂ will be in excess

So, Hydrogen is limiting reactant in this reaction and the amount of NH₃ depends on the amount of hydrogen.

Now

To find mass of NH₃ we will do following calculation

Look at the reaction

As we Know

N₂ + 3 H₂ ------—> 2 NH₃

6 g 2 mol

So, 6 g of hydrogen gives 2 moles of NH₃, then how many moles of NH₃ will be produce by 14.2 g

Apply Unity Formula

6 g of H₂ ≅ 2 mol of NH₃

14.2 g of H₂ ≅ X mol of NH₃

Do cross multiply

X mol of NH₃= 14.2 g x 2 mol / 6 g

X mol of NH₃ = 4.73 mol

So, 14.2 g of hydrogen gives 4.73 moles of NH₃

Now

Convert moles of NH₃ to mass

Formula will be used

mass in grams = no. of moles x molar mass . . . . . . (2)

Molar mass of NH₃

Molar mass of NH₃ = 14 + 3(1)

Molar mass of NH₃ = 14 + 3 = 17 g/mol

Put values in equation 2

mass in grams = 4.73 mole x 17 g/mol

mass in grams = 80.41 g

mass of NH₃= 80.41 g

3 0

2 years ago

If you were to assume a 95% yield for the formation of the phosphonium ion, how many milligrams of benzyl chloride and triphenyl

luda_lava [24]

I am attempting the problem for phosphonium Ion rather than its chloride salt. The chemical equation is shown below along with molar masses in mg.

First of all we will calculate the amounts of reactants required for the synthesis of 220 mg of phophonium ion. Calculations for both reactants is as follow,

For Benzyl chloride,
$\frac{353420 mg of Phosphonium is formed by reacting}{220 mg of Phosphonium will be formed by}$ = $\frac{126580 g of benzyl chloride}{X}$

Solving for X,
X = $\frac{220 mg . 126580 mg}{353420 mg}$
X = 78.79 mg

For PPh₃:
$\frac{353420 mg of Phosphonium is formed by reacting}{220 mg of Phosphonium will be formed by}$ = $\frac{262290 g of PPh3}{X}$

Solving for X,
X = $\frac{220 mg . 262290 mg}{353420 mg}$
X = 163.27 mg

Now, Assuming these values as for 95 % conversion, we can calculate 100 % yield as follow,

when $\frac{95 percent}{100 percent}$ = $\frac{220 g}{X}$
Solving for X,

X = $\frac{220 . 100}{95}$ = 231.57 mg

Now, calculate reactants mass with respect to 231.57 mg
when $\frac{220 mg phosphonium required}{231.57 mg require}$ = $\frac{78.79 g of benzyl chloride}{X}$
Solving for ,

X = $\frac{231.57 . 78.79}{220}$ = 82.93 mg of Benzyl chloride

when $\frac{220 mg phosphonium required}{231.57 mg require}$ = $\frac{163.27 g of PPh3}{X}$
Solving for ,

X = $\frac{231.57 . 163.27}{220}$ = 171.85 mg of PPh3

So, reaction was started with reacting 82.93 mg of Benzyl Chloride and 171.85 mg of Triphenyl Phosphine.

7 0

2 years ago

Which of the following statements is true? (Multiple answers possible) Group of answer choices:

Sophie [7]

Answer:The endpoint does not correspond exactly to the equivalence point

At the endpoint, a change in a physical quantity associated with the equivalence point occurs.

At the equivalence point, the mole number of equivalents of reagent added is equal to the mole number of equivalents of analyte present.

Explanation:

The end point is always indicated by some physical property that changes such as colour. At the equivalence point, the mole number of equivalents of reagent added is equal to the mole number of equivalents of analyte present. The equivalence point cannot be physically observed but can be deduced after a titration curve is plotted.

3 0

2 years ago