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3 years ago

9

Lewis structure for trisulfur

2 answers:

Julli [10]3 years ago

6 0

Answer:

idc

Explanation:

maria [59]3 years ago

5 0

Thank yhu djifjgkfti

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Help pls you get 22 points if you answer me

snow_tiger [21]

Answer:

the 2nd one

Explanation:

All other symbols are for Mathematics.

5 0

2 years ago

Consider the above reaction. If some CO2 were removed from the container, what would the reaction do in an attempt to replace th

faltersainse [42]

Answer:

See explanation

Explanation:

The reaction to be considered is shown below;

H2CO3<------->CO2 + H2O

We know that when a constraint such as a sudden change in concentration, pressure or temperature is imposed on a reaction system in equilibrium, the system has to adjust itself by shifting in a particular direction in order to cancel the constraint.

Now, if we remove CO2, the equilibrium position must shift to the right by the decomposition of more H2CO3 to establish equilibrium again.

4 0

3 years ago

At a certain concentration of H2 and NH3, the initial rate of reaction is 0.120 M / s. What would the initial rate of the reacti

mel-nik [20]

The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

$rate=k[H_2]^2[NH_3]$

At a certain concentration of $H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]$

As we are given that:

Initial rate = 0.120 M/s

Expression for rate law for first observation:

$0.120=k[H_2]^2[NH_3]$ ....(1)

Expression for rate law for second observation:

$R=k(\frac{[H_2]}{2})^2[NH_3]$ ....(2)

Dividing 2 by 1, we get:

$\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}$

$\frac{R}{0.120}=\frac{1}{4}$

$R=0.03M/s$

Therefore, the initial rate of the reaction will be, 0.03 M/s

5 0

3 years ago

Which of the following most likely happens during a physical change?

alexandr1967 [171]

The correct answer is B.

4 0

3 years ago

Read 2 more answers

How many grams of barium sulfate are produced if 25.34 mL of 0.113 M BaCl2 completely react given the reaction: BaCl2 (aq) + Na2

jeyben [28]

<u>Answer:</u> The amount of barium sulfate produced in the given reaction is 0.667 grams.

<u>Explanation:</u>

To calculate the number of moles from molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of barium chloride = 0.113 M

Volume of barium chloride = 25.34 mL = 0.02534 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.113mol/L=\frac{\text{Moles of barium chloride}}{0.02534L}\\\\\text{Moles of barium chloride}=0.00286mol$

For the given chemical reaction:

$BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)$

By Stoichiometry of the reaction:

1 mole of barium chloride is producing 1 mole of barium sulfate.

So, 0.00286 moles of barium chloride will produce = $\frac{1}{1}\times 0.00286mol=0.00286mol$ of barium sulfate.

Now, to calculate the mass of barium sulfate, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of barium sulfate = 233.38 g/mol

Moles of barium sulfate = 0.00286 moles

Putting values in above equation, we get:

$0.00286mol=\frac{\text{Mass of barium sulfate}}{233.38g/mol}\\\\\text{Mass of barium sulfate}=0.667g$

Hence, the amount of barium sulfate produced in the given reaction is 0.667 grams

4 0

3 years ago