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Scilla [17]
3 years ago
9

Please help me on this question I'm confusied on it

Chemistry
2 answers:
Fynjy0 [20]3 years ago
7 0
Find the average speed of the car and enter those numbers.
Korolek [52]3 years ago
3 0

Answer:

So you need to look for the average speed of each car, which would allow you to then enter the numbers.

Explanation:

Hope this helps! Have a great say! :)

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¿Puede ocurrir una reacción química sin generar cambios visibles?
Mrac [35]

Answer:

si

Explanation:

6 0
3 years ago
A child and her mother are at the park. The child walks around the rectangular park while her mom waits on a bench. The child re
RSB [31]
C 1,772 is the answer for the question

7 0
3 years ago
Read 2 more answers
A sample of ammonia gas was allowed to come to equilibrium at 400 K. 2NH3(g) <---> N2(g) 3H2(g) At equilibrium, it was fou
Softa [21]

Answer:

Kc for this equilibrium is 2.30*10⁻⁶

Explanation:

Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of reactants and products are held constant.

Being:

aA + bB ⇔ cC + dD

the equilibrium constant Kc is defined as:

Kc=\frac{[C]^{c}*[D]^{d}  }{[A]^{a} *[B]^{b} }

In other words, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients. Kc is constant for a given temperature, that is to say that as the reaction temperature varies, its value varies.

In this case, being:

2 NH₃(g) ⇔ N₂(g) + 3 H₂(g)

the equilibrium constant Kc is:

Kc=\frac{[N_{2} ]*[H_{2} ]^{3}  }{[NH_{3} ]^{2} }

Being:

  • [N₂]= 0.0551 M
  • [H₂]= 0.0183 M
  • [NH₃]= 0.383 M

and replacing:

Kc=\frac{0.0551*0.0183^{3}  }{0.383^{2} }

you get:

Kc= 2.30*10⁻⁶

<u><em>Kc for this equilibrium is 2.30*10⁻⁶</em></u>

8 0
3 years ago
Thermal energy would naturally flow between objects at which temperatures?
marissa [1.9K]

Answer:

i think the answer is letter C. From 35°c to 45°c

Explanation:

sorry if it is wrong

4 0
3 years ago
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
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