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2 years ago

Why do the air molecules inside a bicycle tire speed up as the temperature gets warmer?

2 answers:

6 0

The correct answer of the given question above would be option A. The air molecules inside a bicycle tire speed up as the temperature gets warmer because the heat is transferred to the molecules and gives them more kinetic energy. <span>When </span>heat<span> is added to a substance, the </span>molecules<span> and atoms vibrate </span>faster<span>. </span>

Mice21 [21]2 years ago

5 0

Answer: Option (a) is the correct answer.

Explanation:

When temperature of surrounding atmosphere around the tire increases then it will lead to increase in the kinetic energy of molecules.

Due to increase in kinetic energy of molecules there will be more number of collisions between the air molecules that are present inside the bicycle tire.

Hence, it means heat has been transferred from the surrounding atmosphere to the molecules inside the tire.

Thus, we can conclude that the air molecules inside a bicycle tire speed up as the temperature gets warmer because the heat is transferred to the molecules and gives them more kinetic energy.

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olasank [31]

Answer:

Explanation:

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2 years ago

Help in chem please!!!!!!

IgorLugansk [536]

This is a one-step unit analysis problem. Since we are staying in moles, grams of our compound, and thus molar mass, is not needed.

1 mole is equal to 6.022x10²³ particles as given, so:

$1.5x10^{24} particles FeO_{2} (\frac{1mol}{6.022x10^{23}particles } )=2.49 molFeO_{2}$

<h3>Answer:</h3>

2.49 mol

Let me know if you have any questions.

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2 years ago

Radioactive radium has a half-life of approximately 1,599 years. The initial quantity is 13 grams. How much (in grams) remains a

Luda [366]

The quantity of substance remains after 850 years is 8.98g if the half life of radioactive radium is 1,599 years.

<h3>What is half life period? </h3>

The time taken by substance to reduce to its half of its initial concentration is called half life period.

We will use the half- life equation N(t)

N e^{(-0.693t) /t½}

Where,

N is the initial sample

t½ is the half life time period of the substance

t2 is the time in years.

N(t) is the reminder quantity after t years .

Given

N = 13g

t = 350 years

t½ = 1599 years

By substituting all the value, we get

N(t) = 13e^(0.693 × 50) / (1599)

= 13e^(- 0.368386)

= 13 × 0.691

= 8.98

Thus, we calculated that the quantity of substance remains after 850 years is 8.98g if the half life of radioactive radium is 1,599 years.

learn more about half life period:

brainly.com/question/20309144

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1 year ago

Planes X and Y and points C, D, E, and F are shown.

Citrus2011 [14]

Answer:

The line drawn through points D and E.

Explanation:

Y is a horizontal line and D and E are both on the same line. If a line were drawn it would be within the Y plane.

3 0

2 years ago

Read 2 more answers

At 25 °C, the equilibrium partial pressures for the following reaction were found to be PA = 8.00 atm, PB = 5.40 atm, PC = 6.90

Yuliya22 [10]

<u>Answer:</u> The standard change in Gibbs free energy for the given reaction is 4.33 kJ/mol

<u>Explanation:</u>

For the given chemical equation:

$A(g)+2B(g)\rightleftharpoons C(g)+D(g)$

The expression of $K_p$ for the given reaction:

$K_p=\frac{p_C\times p_D}{p_A\times (p_B)^2}$

We are given:

$p_A=8.00atm\\p_B=5.40atm\\p_C=6.90atm\\p_D=5.90atm$

Putting values in above equation, we get:

$K_p=\frac{6.90\times 5.90}{8.00\times (5.40)^2}\\\\K_p=0.174$

To calculate the standard Gibbs free energy, we use the relation:

$\Delta G^o=-RT\ln K_p$

where,

$\Delta G^o$ = standard Gibbs free energy

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 0.174

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/Kmol)\times 298K\times \ln (0.174)\\\\\Delta G^o=4332.5J/mol=4.33kJ/mol$

Hence, the standard change in Gibbs free energy for the given reaction is 4.33 kJ/mol

3 0

3 years ago