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viktelen [127]
3 years ago
9

Given two temperatures, 36°F and 72°F, which of the following correctly compares the number of opportunities for particles to al

ign during a chemical reaction?
A) same at both B) greater at 36°F C) greater at 72°F D) does not depend on temperature?
Chemistry
2 answers:
bogdanovich [222]3 years ago
5 0

Answer:

C) greater at 72°F

Explanation:

Hello,

Here, it is important to remember that when a chemical reaction commences, the reactants' molecules start to crash to yield the new species, the products. In such a way, when the temperature increases, such crashes become stronger and more repetitive as the molecules gain more energy to move faster and consequently ease the chemical change. Such crashes are directly related with the number of opportunities, say probabilities, the molecules have to crash and subsequently form the products.

Best regards.

Ivanshal [37]3 years ago
3 0
The answer is C) greater at 72 °F


hope it helped
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MissTica
Okay I did the math and I'm guessing around 18*C
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Convert 5° C into Fahrenheit.
Lady_Fox [76]

Answer:

5 in celcius is 41 in fahrenheit!

Explanation:

7 0
3 years ago
A 700.0 mL gas sample at STP is compressed to a volume of 200.0 mL, and the temperature is increased to 30.0°C. What is the new
dolphi86 [110]

Answer: The new pressure of the gas in Pa is 388462

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas at STP = 10^5Pa

P_2 = final pressure of gas = ?

V_1 = initial volume of gas = 700.0 ml

V_2 = final volume of gas = 200.0 ml

T_1 = initial temperature of gas = 273 K

T_2 = final temperature of gas = 30^oC=273+30=303K

Now put all the given values in the above equation, we get:

\frac{10^5\times 700.0ml}{273K}=\frac{P_2\times 200.0ml}{303K}

P_2=388462Pa

The new pressure of the gas in Pa is 388462

6 0
3 years ago
__Zn(s)+__CuSO4(aq)--->_______+________
stiks02 [169]

Answer:

Zn + CuSO4 —> ZnSO4 + Cu

Explanation:

Zn is higher than Cu in electrochemical series and so will displaces Cu in solution according to the equation:

Zn + CuSO4 —> ZnSO4 + Cu

5 0
3 years ago
Calculate the volume in mL of 0.279 M Ca(OH)2 needed to neutralize 24.5 mL of 0.390 M H3PO4 in a titration.
Vsevolod [243]

The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL

<h3>Balanced equation </h3>

2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O

From the balanced equation above,

  • The mole ratio of the acid, H₃PO₄ (nA) = 2
  • The mole ratio of the base, Ca(OH)₂ (nB) = 3

<h3>How to determine the volume of Ca(OH)₂ </h3>
  • Molarity of acid, H₃PO₄ (Ma) = 0.390 M
  • Volume of acid, H₃PO₄ (Va) = 24.5 mL
  • Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
  • Volume of base, Ca(OH)₂ (Vb) =?

MaVa / MbVb = nA / nB

(0.39 × 24.5) / (0.279 × Vb) = 2/3

9.555 / (0.279 × Vb) = 2/3

Cross multiply

2 × 0.279 × Vb = 9.555 × 3

0.558 × Vb = 28.665

Divide both side by 0.558

Vb = 28.665 / 0.558

Vb = 51.4 mL

Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL

Learn more about titration:

brainly.com/question/14356286

5 0
2 years ago
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