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Semenov [28]
3 years ago
13

If 34.7 g of AgNO₃ react with 28.6 g of H₂SO₄ according to this UNBALANCED equation below, how many grams of Ag₂SO₄ could be for

med? AgNO₃(aq) + H₂SO₄ (aq) → Ag₂SO₄ (s) + HNO₃ (aq)
Chemistry
1 answer:
Luba_88 [7]3 years ago
3 0

The  number  of grams   of Ag2SO4  that could be formed  is   31.8  grams



    <u><em> calculation</em></u>

Balanced   equation is  as below

2 AgNO3 (aq)  + H2SO4(aq)  →  Ag2SO4 (s)   +2 HNO3 (aq)


  • Find  the  moles  of  each reactant by use  of  mole= mass/molar mass  formula

that is  moles of  AgNO3= 34.7 g / 169.87  g/mol= 0.204 moles

             moles of  H2SO4 =  28.6  g/98  g/mol  =0.292  moles

  • use the  mole  ratio to determine the moles of  Ag2SO4

   that is;

  •    the mole ratio of  AgNo3 : Ag2SO4 is  2:1 therefore  the  moles of Ag2SO4=  0.204  x1/2=0.102 moles

  • The moles  ratio of H2SO4  : Ag2SO4  is  1:1  therefore  the moles of Ag2SO4 = 0.292  moles

 

  •      AgNO3  is the limiting reagent therefore  the moles of   Ag2SO4 = 0.102  moles

<h3>     finally  find  the mass  of Ag2SO4  by use of    mass=mole  x molar mass  formula</h3>

that  is  0.102   moles  x  311.8  g/mol= 31.8 grams

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fredd [130]

A hypothesis is an experimental question that can later be proven correct or incorrect by scientific evidence.

7 0
3 years ago
Naturally occurring element X exists in three isotopic forms: X-28 (27.979 amu, 92.21% abundance), X-29 (28.976 amu 4.70% abunda
bezimeni [28]

<u>Answer:</u> The average atomic mass of X is 28.09 amu

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i .....(1)

  • <u>For isotope 1:</u>

Mass of isotope 1 = 27.979 amu

Percentage abundance of isotope 1 = 92.21 %

Fractional abundance of isotope 1 = 0.9212

  • <u>For isotope 2:</u>

Mass of isotope 2 = 28.976 amu

Percentage abundance of isotope 2 = 4.70 %

Fractional abundance of isotope 2 = 0.0470

  • <u>For isotope 3:</u>

Mass of isotope 3 = 29.974 amu

Percentage abundance of isotope 3 = 3.09 %

Fractional abundance of isotope 3 = 0.0309

Putting values in equation 1, we get:

\text{Average atomic mass of X}=[(27.979\times 0.9212)+(28.976\times 0.0470)+(29.974\times 0.0309)]

\text{Average atomic mass of X}=28.09amu

Hence, the average atomic mass of X is 28.09 amu

4 0
3 years ago
Solid potassium chlorate decomposes upon heating to form
olga55 [171]

Answer:

32.6%

Explanation:

Equation of reaction

2KClO₃ (s) → 2KCl (s) + 3O₂ (g)

Molar mass of 2KClO₃ = 245.2 g/mol ( 122.6 × 2)

Molar volume of Oxygen at s.t.p = 22.4L / mol

since the gas was collected over water,

total pressure = pressure of water vapor + pressure of  oxygen gas

0.976 = 0.04184211 atm + pressure of oxygen gas at 30°C

pressure of oxygen = 0.976 - 0.04184211 = 0.9341579 atm = P1

P2 = 1 atm, V1 = 789ml, V2 = unknown, T1 = 303K, T2 = 273k at s.t.p

Using ideal gas equation

\frac{P1V1}{T1} = \frac{P2V2}{T2}

V2 = \frac{P1V1T2}{T1P2}

V2 = 664.1052 ml

245.2   yielded 67.2 molar volume of oxygen

0.66411 will yield = \frac{245.2 * 0.66411}{67.2}  = 2.4232 g

percentage of potassium chlorate in the original mixture = \frac{2.4232 * 100}{7.44} = 32.6%

3 0
3 years ago
Sixty liters of a gas were collected over water when the barometer read 663 mmhg , and the temperature was 20∘c. what volume wou
lesya [120]
First, let's compute the number of moles in the system assuming ideal gas behavior. 

PV = nRT
(663 mmHg)(1atm/760 mmHg)(60 L) = n(0.0821 L-atm/mol-K)(20+273 K)
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n = 2.176 moles

At standard conditions, the standard molar volume is 22.4 L/mol. Thus,

Standard volume = 22.4 L/mol * 2.176 mol =<em> 48.74 L</em>
3 0
3 years ago
What is the percent by volume of isopropyl alcohol in a solution that contains 45 ml of isopropyl alcohol in 1.4 L of water?
11Alexandr11 [23.1K]

Answer:

3.11%

Explanation:

We express a solution's volume by volume percent concentration, % v/v,

Take the ratio of the isopropyl alcohol (IPA) volume to the total volume of the solution, which is 1400 mL of water+ 45 mL of IPA,

and multiply by 100 to get the percentage:    45/(1400+45) *100 = 0.0311*100 =  3.11%

6 0
2 years ago
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