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I am Lyosha [343]
3 years ago
11

At stp, 1 l of br2 reacts with 3 l of f2 to completion and produces 2l of a product. what is the formula of the product

Chemistry
1 answer:
vredina [299]3 years ago
4 0
The formula of the product   when Br2  reacts with 3L  of F2 to   completion and produce 2 L of the product is

       2 BrF3
Br2  +3F2 = 2BrF3

1 mole  of  Br  reacted  3 moles of f2 to form 2  moles of BrF3
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C2F4 effuses through a barrier at a rate of 4.6x10-6 mol/hour, while an unknown gas effuses at a rate of 5.8x10-6 mol/hour. What
umka21 [38]
The  molar mass  of  the unknown  compound  is   calculated   as   follows

let the unknown  gas be represented by   letter  Y

Rate of C2F4/  rate of  Y  = sqrt of   molar  mass of gas Y/ molar mass of  C2F4

 =  (4.6  x10^-6/ 5.8  x10^-6)  = sqrt  of  Y/ 100

remove  the  square  root  sign  by  squaring  in both  side

(4.6  x  10^-6 / 5.8  x10^-6)^2 =  Y/100

= 0.629 =Y/100

multiply  both side  by  100

Y=  62.9 is  the molar  mass of unknown  gas



5 0
3 years ago
Given 4.80g of ammonium carbonate, find:
V125BC [204]

Answer:

1) 0.05 mol.

2) 0.1 mol.

3) 0.05 mol.

4) 0.4 mol.

5) 2.4 x 10²³ molecules.

Explanation:

<em>1) Number of moles of the compound:</em>

no. of moles of ammonium carbonate = mass/molar mass = (4.80 g)/(96.09 g/mol) = 0.05 mol.

<em>2) Number of moles of ammonium ions :</em>

  • Ammonium carbonate is dissociated according to the balanced equation:

<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

<em>∴ The no. of moles of NH₄⁺ ions in 0.05 mol of (NH₄)₂CO₃ </em>= (2.0)(0.05 mol) =  <em>0.1 mol.</em>

<em>3) Number of moles of carbonate ions :</em>

  • Ammonium carbonate is dissociated according to the balanced equation:

<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of CO₃²⁻ ions in 0.05 mol of (NH₄)₂CO₃ = (1.0)(0.05 mol) = 0.05 mol.

<em>4) Number of moles of hydrogen atoms:</em>

  • Every 1.0 mol of (NH₄)₂CO₃  contains:

2.0 moles of N atoms, 8.0 moles of H atoms, 1.0 mole of C atoms, and 3.0 moles of O atoms.

<em>∴ The no. of moles of H atoms in 0.05 mol of (NH₄)₂CO</em>₃ = (8.0)(0.05 mol) = <em>0.4 mol.</em>

<em>5) Number of hydrogen atoms:</em>

  • It is known that every mole of a molecule or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.

<u><em>Using cross multiplication:</em></u>

1.0 mole of H atoms contains → 6.022 x 10²³ atoms.

0.4 mole of H atoms contains → ??? atoms.

<em>∴ The no. of atoms in  0.4 mol of H atoms</em> = (6.022 x 10²³ molecules)(0.4 mole)/(1.0 mole) = <em>2.4 x 10²³ molecules.</em>

8 0
3 years ago
The molar madd of Ca(OH)2
Nonamiya [84]
What you have to do is find a periodic table and add the mass of each atom that the compound is made of.

Ca= 40.1
O= 16.0
H= 1.01

keep in mind that you have to also account for how many atoms of each there are in the molecule. for example, in Ca(OH)2, there are one Ca, two O and two H

so the molar mass of Ca(OH)2= 40.1 + (2 x 16.0) + (2 x 1.01)= 74.12 g/mol
8 0
3 years ago
Please help. It's for chemistry
irinina [24]
What do i help with what
3 0
2 years ago
A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the
horrorfan [7]

Answer : (C) Hafnium is the most likely identity of the given substance.

Solution :  Given,

Mass of given substance (m) = 46.9 g

Volume of given substance (V) = 3.5 Cm^{3}

First, find the Density of given substance.

Formula used :    

Density=\frac{\text{Mass of given substance}}{\text{Voume of given substance}}

Now,put all the values in this formula, we get

Density=\frac{46.9 g}{3.5 Cm^{3} } = 13.4 g/Cm^{3}

So, we conclude that the density of given substance (13.4 g/Cm^{3}) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/Cm^{3} respectively).

According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.

Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.

So, Hafnium is the most likely element which is the identity of the given substance.

3 0
3 years ago
Read 2 more answers
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