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Roman55 [17]
3 years ago
5

Consider the following reaction:

Chemistry
1 answer:
rusak2 [61]3 years ago
8 0

Answer:

1. a) 0,18

2. e) The reaction proceeds to the left, forming more BrCl(g).

Explanation:

1. In a gas reaction as:

2SO₂(g) +O₂(g) ⇄ 2SO₃(g)

it is possible to convert kp to kc using:

kp = kc (RT)^Δn

Where kp is gas equilibrium constant, kc is equilibrium constant (13), R is gas constant (0,082atmL/molK), T is temperature (900K), and Δn is number of moles of gas products - number of moles of gas reactants (That is 2 - (2+1) = -1). Replacing:

kp = 13×(0,082atmL/molK×900K)^-1

<em>kp = 0,18</em>

<em></em>

2. Based on Le Chatelier's principle, the change in temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system. For the reaction:

2 BrCl(g) ⇄ Br₂(g) + Cl₂(g).

The addition of 0,050M of each compound cause <em>the reaction proceeds to the left, forming more BrCl(g)</em> because based on the reaction, you need two moles of BrCl per mole of Br₂(g) and Cl₂(g) to keep the system in the same. But you are adding the same proportion of moles of each compound.

I hope it helps!

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The solubility of magnesium phosphate at a given temperature is 0.173 g/L. Calculate the Ksp at this temperature. After you calc
jek_recluse [69]

Answer: K_{sp}=1.25\times 10^{-14}

pK_{sp}=13.90

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as K_{sp}  

The equation for the ionization of magnesium phosphate is given as:

Mg_3(PO_4)_2\rightarrow 3Mg^{2+}+2PO_4^{3-}

 When the solubility of Mg_3(PO_4)_2 is S moles/liter, then the solubility of Mg^{2+} will be 3S moles\liter and solubility of PO_4^{3-} will be 2S moles/liter.

Thus S = 0.173 g/L or \frac{0.173g/L}{262.8g/mol}=0.00065mol/L

K_{sp}=(3S)^3\times (2S)^2

K_{sp}=108S^5

K_{sp}=108\times (0.00065)^5=1.25\times 10^{-14}

pK_{sp}=-log(K_{sp})=\log (1.25\times 10^{-14})=13.90

5 0
2 years ago
A sample of gas has a volume of 20.0 mL at STP. What will the volume be if the temperature is changed to 546 K and the pressure
Ostrovityanka [42]

The volume did not change, it remained at 20 ml

<h3>Further explanation</h3>

Given

20 ml a sample gas at STP(273 K, 1 atm)

T₂=546 K

P₂=2 atm

Required

The volume

Solution

Combined gas Law :

\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}

Input the value :

\tt \dfrac{1\times 20}{273}=\dfrac{2\times V_2}{546}\\\\V_2=\dfrac{1\times 20\times 546}{273\times 2}\\\\V_2=20~ml

The volume does not change because the pressure and temperature are increased by the same ratio as the initial conditions (to 2x)

5 0
2 years ago
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OlgaM077 [116]

I think the answer is yes

3 0
2 years ago
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harkovskaia [24]
A because that honestly just makes the most sense
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If 2.03 g of oxygen react with carbon monoxide, how many grams of CO2 are formed?
mamaluj [8]

Answer:

The suitable equation for this reaction is

2CO + O₂ -----> 2CO₂

Here, we are given that we have 2 grams of O₂

From the equation, we can see that 2 * Moles of O₂ = Moles of CO₂

Moles of O₂:

2/32 = 1/16 moles

Therefore, the number of moles of CO₂ is twice the moles of O₂

Moles of CO₂ = 2 * 1/16

Moles of CO₂  formed = 1/8 moles

Mass of CO₂  formed = Molar mass of CO₂ * Moles of CO₂

Mass of CO₂  formed = 44 * 1/8

Mass of CO₂  formed = 5.5 grams

Hence, option B is correct

Kindly Mark Brainliest, Thanks!!!

8 0
3 years ago
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