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dimulka [17.4K]
2 years ago
8

Hydrogen gas and bromine gas react to form hydrogen bromide gas. How much heat (kJ) is released when 155 grams of HBr is formed

in this reaction? ΔH° = -72 kJ.
Chemistry
1 answer:
VashaNatasha [74]2 years ago
8 0

Answer:

-37.63KJ

Explanation:

First, we write an equation of reaction:

H + Br ——-> HBr

Now we are told that 155g of HBr is formed. We can calculate the number of moles of HBr formed as follows:

We simply divide the mass of the HBr formed by the molar mass of HBr. The molar mass of HBr is 81g/mol

The number of moles of HBr formed is thus 81/155 = 0.52 moles

The amount of heat released is thus 0.52 * (-72) = -37.63KJ

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bija089 [108]

Ksp of AgCl= 1.6×10⁻¹⁰

AgCl=Ag⁺ +Cl⁻

Ksp=[Ag⁺][Cl⁻]

Assume [Ag⁺]=[Cl⁻]=x

Ksp=x²

1.6×10⁻¹⁰=x²

x=0.000012

In FeCl₃:

FeCl₃------>Fe⁺³+ 3Cl⁻

as there is 0.010 M FeCl₃

So there will be ,

[Cl⁻]= 0.030

So

[Ag⁺]=Ksp/[Cl⁻]

=1.6×10⁻¹⁰/0.030

=5.3×10⁻⁹

so solubility of AgCl in FeCl₃ will be 5.3×10⁻⁹.

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2 years ago
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1. The specific heat capacity of iron is 0.461 J g–1 K–1 and that of titanium is 0.544 J g–1 K–1. A sample consisting of a mixtu
mezya [45]

Answer:

The answer is 80,1 °C

Explanation:

Let´s start from the mass of the sample and the heat capacities:

First of all, we must calculate an average heat capacity. That's because we have a mixture and it is unknown the heat capacity of the whole sample.

The way we should do this calculation is as follows:

(1) H_{average}=Mass Fraction_{first component}* H_{first component}+MassFraction_{secondcomponent}*H_{second component}

For example, the mass fraction of Fe is simply:

(2) MassFraction_{Fe}=\frac{10g Fe}{10g Fe + 10gTi}=0.5

If you combine the equations (1) and (2) you have:

(3) H_{average}=0.5*0.461+0.5*0.544=0.5025\frac{J}{g-K}

Once calculated the average heat capacity we can solve the problem taking into account the corresponding equation:

(4) Q=m*H_{average}*(T_2-T_1)

Remember that:

<em>Q:</em> Heat gained or lost

<em>m</em>: Mass of the sample you want to analize

H_{average} : The value obtained in equation (3)

T_2: Final temperature of the sample

T_1: Initial temperature of the sample

Now we must replace the problem data in equation (4)

Take into account:

  • Heat gained in a system have a positive value
  • Heat lost in a system have a negative value
  • In this problem the sample loses 200 J, for this reason Q=-200J
  • The mass of the whole sample is: 10g of Fe + 10g of Ti = 20g of sample
  • The temperatures must be in absolute units of temperature (these are: rankine or kelvin)
  • The initial temperature of the system is 100°C or 373K

Now we are ready to use equation (4):

(5) -200J=20g*0.5025\frac{J}{g*K} *(T_2-373K)

It is clear that the unknown in equation (5) is T_2

The next step is to calculate T_2. Don't forget the signs; these are important.

Key concept: <u>Since the system is loosing heat, the final temperature of the system ( T_2) should be lower than the initial temperature ( T_1 )</u>

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Sound _____.
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The answer is the first one that say <span>Travels in longitudinal waves</span>
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A rock displaces 1,500 mL of water. The volume of the rock is
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1500 cm^3 ; 1 mL equals 1 cm^3
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a sample of an oxide of iron was reduced to iron by heating with hydrogen. the mass of iron obtained was 4.35g and mass of water
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Answer:

FeO(s) + H2(g)→ Fe(s) + H2O(g)

Explanation:

Moles of Iron will be = 4.35 g/55.845 g/mol

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The mole ratio of Iron to water

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= 1   :  1.3

= 1 : 1

Therefore, the equation for the reaction is;

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