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scoundrel [369]
4 years ago
9

What mass of hbr (in g) would you need to dissolve a 3.4 −g pure iron bar on a padlock?

Chemistry
1 answer:
vivado [14]4 years ago
5 0

<u>Given:</u>

Mass of pure iron (Fe) = 3.4 g

<u>To determine:</u>

Mass of HBr needed to dissolve the above iron

<u>Explanation:</u>

Reaction between HBr and Fe is

Fe + 2HBr → FeBr₂ + H₂

Based on the reaction stoichiometry-

1 mole of Fe reacts with 2 moles of HBr

# moles of Fe = mass of Fe/atomic mass of Fe = 3.4/56 g.mol⁻¹ = 0.0607 moles

Therefore # moles of HBr = 2*0.0607 = 0.1214 moles

Molar mass of HBr = 81 g/mole

Mass of HBr = 0.1214 moles * 81 g/mole = 9.83 g

Ans: Mass of HBR required is 9.83 g

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Complete question:

Which of the following situations contains an example of a chemical reaction?

a. Ice forming after water is placed in a freezer

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6 0
1 year ago
What is the most corrosive element in the periodic table?
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4 0
3 years ago
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Use the molar heat capacity for aluminum from table 1 to calculate the amount of energy needed to raise the temperature of 260.5
Nimfa-mama [501]
Unfortunately, you failed to include the table 1 from which the molar heat capacity of aluminum could have been obtained. However, as a general rule, the heat needed to raise the temperature of a certain substance by certain degrees is calculated through the equation,
                            H = mcpdT
where H is heat, m is mass, cp is specific heat capacity, and dT is change in temperature. From a reliable source, cp for aluminum is equal to 0.215 cal/g°C. Substituting this to the equation,
                               H = (260.5 g)(0.215 cal/g°C)(125°C - 0)
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6 0
3 years ago
Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 1
NeTakaya

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x                       x               x

The dissociation constant from the above reactions is given by:

Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}

6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}

6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75    

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

\% = \frac{x}{[HCN]} \times 100

\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100

\% = 3.5 \cdot 10^{-3} \%          

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!      

6 0
3 years ago
For the reaction so3 + h2o h2so4, calculate the percent yield if 500. g of sulfur trioxide react with excess water to produce 57
Lelu [443]
The %  yield  if  500 g of  sulfur trioxide  reacted  with  excess  water to   produce  575 g  of  sulfuric  acid is calculated using  the  below  formula


%  yield = actual  yield/ theoretical  yield  x100

actual  yield =575 grams
to  calculate  theoretical  yield
find the  moles  of SO3   used =mass/molar  mass
=  500g/   80 g/mol =6.25  moles

SO3+H2O=H2SO4
by   use of  mole ratio  of SO3  :  H2SO4 which  is 1:1  the moles of H2SO4  is  also=  6.25  moles

the theoretical  yield of H2SO4 is therefore =  moles /molar  mass
=  6.25  x98=  612.5 grams

%yield  is therefore= 575 g/612 g   x100=  93.9  %

5 0
3 years ago
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