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VladimirAG [237]
3 years ago
15

Decomposition of NI3 produces nitrogen gas and iodine. How many grams of the reactant would be required to react completely to g

ive 7.97g of the iodine.
Chemistry
1 answer:
VMariaS [17]3 years ago
5 0

Answer:

Grams of the reactant would be required is 8.26 g

Explanation:

Decomposition of NI3 is :

2 NI3 (s) --------------------> N2 (g)............... +..................... 3 I2 (g)

2 mol..................................1 mol.........................................3 mol

2 x 394.7 g/mol..............28 g/mol....................................3 x 253.8 g/mol

789.4 g...............................28 g...........................................761.4 g

From chemical equation it is cleared that

761.4 g of iodine is obtained from = 789.4 g of NI3

Therefore,

7.97 g of iodine is obtained from = 789.4 g x 7.97 g / 761.4 of NI3

                                                       = 6291.518 / 761.4 of NI3

                                                       =  8.26 g  of NI3

Hence

Grams of the reactant would be required = 8.26 g

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If the initial concentration of SO2Cl2 is 0.125 M , what is the concentration of SO2Cl2 after 210 s ?
andre [41]

Complete question:

The decomposition of SO2Cl2 is first order in SO2Cl2 and has a rate constant of 1.44×10⁻⁴ s⁻¹ at a certain temperature.

If the initial concentration of SO2Cl2 is 0.125 M , what is the concentration of SO2Cl2 after 210 s ?

Answer:

After 210 s the concentration of SO2Cl2  will be 0.121 M

Explanation:

ln\frac{[A_t]}{[A_0]} =-kt

where;

At is the concentration of A at a time t

A₀ is the initial concentration of A

k is rate constant = 1.44×10⁻⁴ s⁻¹

t is time

ln(At/A₀) = -( 1.44×10⁻⁴)t

ln(At/0.125) =  -( 1.44×10⁻⁴)210

ln(At/0.125) = -0.03024

\frac{A_t}{0.125} = e^{-0.03024

At/0.125 = 0.9702

At = 0.125*0.9702

At = 0.121 M

Therefore, after 210 s the concentration of SO2Cl2  will be 0.121 M

6 0
3 years ago
HELLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLPPPPPPPPPPPPPPPPPPPPPPP
guajiro [1.7K]
Hiii,so the problem is ‍♀️
5 0
3 years ago
consider the reaction between calcium oxide and carbon dioxide: cao ( s ) + co 2 ( g ) → caco 3 ( s ) a chemist allows 14.4 g of
MakcuM [25]

Answer:

CaO is the limiting reagent

Theoritical yield = 25.71 g

% Yield = 75.44%

Explanation:

1 mole = Molar mass of the substance

Molar Mass of CaO = 56 g/mol

Molar Mass of CaCO3 = 100 g/mol

Molar mass of CO2 = 44 g/mol

The balanced Equation is :

CaO + CO_{2}\rightarrow CaCO_{3}

1 mole of CaO reacts with = 1 mole of CO2

56 g of CaO reacts with = 44 g of CO2

1 g of CaO reacts with =

\frac{44}{56}

= 0.785 g of CO2

So,

<u>14.4 g of CaO</u><u> </u>must react with = (14.4 x 0.785) g of CO2

= 11.31 g of CO2

<u>Needed = 11.31 g</u>

<u>Available CO2  = 13.8 g</u><u> </u>(given)

So CO2 is in excess , hence<u> CaO is the limiting reagent and product will produce from 14.4 g of CaO</u>

CaO + CO_{2}\rightarrow CaCO_{3}

1 mole of CaO will produce 1 mole pf CaCO3

56 g of CaO produce = 100 g of CaCO3

1 g  of CaO produce =

\frac{100}{56}

= 1.785 g of CaCO3

14.4 g of CaO will produce = (1.785 x 14.4) g of CaCO3

= 25.71 g of CaCO3

Theoritical Yield of CaCO3 = 25.71 g

Actual yield = 19.4 g

Percent Yield =

\frac{Actual\ yield}{Theoritical\ yield}\times 100

\frac{19.4}{25.71}\times 100

= 75.44 %

6 0
3 years ago
In an atom of an element there are a fixed number of electrons. If the total mass of electrons in 1 mole (6.022 x 1023) of atoms
prohojiy [21]

Answer:

No. of protons = 34

Explanation:

First we need to calculate the number of electrons in one mole of the the element:

No. of electrons per mole = Total Mass of Electrons/Mass of 1 Electron

No. of electrons per mole = (18.65 x 10⁻³ g)/(9.109 x 10⁻²⁸ g)

No. of electrons per mole = 2.04 x 10²⁵ electrons/mol

Now, we calculate the no. of electrons in 1 atom:

No. of electrons per atom = No. of Electrons per mole/No. of atoms per mole

No. of electrons per atom = (2.04 x 10²⁵ electrons/mol)/(6.022 x 10²³ atoms/mol)

No. of electrons per atom = 34 electrons/atom

Since, the no. of protons in a pure element are equal to the number of electrons. Therefore,

<u>No. of protons = 34</u>

5 0
3 years ago
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Answer:

Sound cannot travel through vacuum

Explanation:

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3 years ago
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