Which of the following is NOT a sign (indicator) that a chemical reaction has occurred?

in 1990, the men's singles winner of the us. open tennis tournament has his serves clocked at 127 mi/hr. How fast must a 56.6 g

Naddika [18.5K]

This problem is providing information about the mass of a tennis ball, 56.6 g (0.0566 kg) and asks for the velocity it will have to equal the wavelength of green light, which is 5400 A or 540 nm (5.4x10⁻⁷ m). Thus, after doing the math, the result is 2.17x10⁻²⁶ m/s.

<h3>Broglie's wavelength:</h3>

In this case, we recall the formula of the Broglie's wavelength as shown below:

$\lambda =\frac{h}{mv}$

Whereas lambda is the wavelength, h is the Planck's constant, m the mass and v the speed; thus, we solve for the speed according to the question:

$v =\frac{h}{m\lambda}$

<h3>Calculations:</h3>

Then, we just plug in the numbers we were given to get the answer:

$v =\frac{6.626x10^{-34} kg*\frac{m^2}{s} }{0.056kg*5.4x10^{-7}m}\\\\v=2.17x10^{-26}m/s$

Learn more about Broglie's wavelength: brainly.com/question/5440536

3 0

2 years ago

Be sure to answer all parts. The equilibrium constant (Kp) for the reaction below is 4.40 at 2000. K. H2(g) + CO2(g) ⇌ H2O(g) +

nikklg [1K]

<u>Answer:</u>

<u>For 1:</u> The value of $\Delta G$ for the chemical equation is -24.636 kJ/mol

<u>For 2:</u> The value of $\Delta G$ for the chemical equation is -20.925 kJ/mol

<u>Explanation:</u>

For the given chemical equation:

$H_2(g)+CO_2(g)\rightleftharpoons H_2O(g)+CO(g)$

<u>For 1:</u>

To calculate the $\Delta G$ for given value of equilibrium constant, we use the relation:

$\Delta G=-RT\ln K_p$ .....(1)

where,

$\Delta G$ = ? kJ/mol

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

T = temperature = 2000 K

$K_p$ = equilibrium constant in terms of partial pressure = 4.40

Putting values in above equation, we get:

$\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (4.40)\\\\\Delta G=-24636.12J/mol$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -24636.12 J/mol = -24.636 kJ/mol

Hence, the value of $\Delta G$ for the chemical equation is -24.636 kJ/mol

<u>For 2:</u>

The expression of $K_p$ for the given chemical equation is:

$K_p=\frac{p_{CO}p_{H_2O}}{p_{H_2}p_{CO_2}}$

We are given:

$p_{CO}=1.18atm\\p_{H_2O}=0.66atm\\p_{CO_2}=0.82atm\\p_{H_2}=0.27atm$

Putting values in above equation, we get:

$K_p=\frac{1.18\times 0.66}{0.27\times 0.82}\\\\K_p=3.52$

Now, calculating the value of $\Delta G$ by using equation 1:

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

T = temperature = 2000 K

$K_p$ = equilibrium constant in terms of partial pressure = 3.52

Putting values in equation 1, we get:

$\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (3.52)\\\\\Delta G=-20925.68J/mol$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -20925.68 J/mol = -20.925 kJ/mol

Hence, the value of $\Delta G$ for the chemical equation is -20.925 kJ/mol

3 0

3 years ago

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The Answer To This Is B.

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