All categories

3 years ago

1. Calculate the energy of a photon with a frequency of 3x1015 Hz. ?????

Chemistry

1 answer:

Afina-wow [57]3 years ago

6 0

Answer:

1.99 x 10⁻¹⁸J

Explanation:

Given parameters:

Frequency of the wave = 3 x 10¹⁵Hz

Unknown:

Energy of the photon = ?

Solution:

To solve this problem, we use the expression below;

E = hf

Where E is the energy, h is the Planck's constant and f is the frequency

Now insert the parameters and solve for E;

E = 6.63 x 10⁻³⁴ x 3 x 10¹⁵ = 19.9 x 10⁻¹⁹J or 1.99 x 10⁻¹⁸J

You might be interested in

Q5 what's the answer to A?

Snezhnost [94]

The answer is CH3CH2CH2CHO

8 0

2 years ago

A 1.25 g gas sample occupies 663 ml at 25∘ c and 1.00 atm. what is the molar mass of the gas?

lakkis [162]

Using ideal gas equation,

$P\times V=n\times R\times T$

Here,

P denotes pressure

V denotes volume

n denotes number of moles of gas

R denotes gas constant

T denotes temperature

The values at STP will be:

P=1 atm

T=25 C+273 K =298.15K

V=663 ml=0.663L

R=0.0821 atm L mol ⁻¹

Mass of gas given=1.25 g g

Molar mass of gas given=?

$Number of moles of gas, n= \frac{Given mass of the gas}{Molar mass of the gas}$

$Number of moles of gas, n= \frac{1.25}{Molar mass of the gas}$

Putting all the values in the above equation,

$1\times 0.663=\frac{1.25}{Molar mass of the gas}\times 0.0821\times 298.15$

Molar mass of the gas=46.15

3 0

2 years ago

Can anybody check my answer?

anzhelika [568]

Answer:

$\boxed{\text{25. 20 L; 26. 49 K}}$

Explanation:

25. Boyle's Law

The temperature and amount of gas are constant, so we can use Boyle’s Law.

$p_{1}V_{1} = p_{2}V_{2}$

Data:

$\begin{array}{rcrrcl}p_{1}& =& \text{100 kPa}\qquad & V_{1} &= & \text{10.00 L} \\p_{2}& =& \text{50 kPa}\qquad & V_{2} &= & ?\\\end{array}$

Calculations:

$\begin{array}{rcl}100 \times 10.00 & =& 50V_{2}\\1000 & = & 50V_{2}\\V_{2} & = &\textbf{20 L}\\\end{array}\\\text{The new volume will be } \boxed{\textbf{20 L}}$

26. Ideal Gas Law

We have p, V and n, so we can use the Ideal Gas Law to calculate the volume.

pV = nRT

Data:

p = 101.3 kPa

V = 20 L

n = 5 mol

R = 8.314 kPa·L·K⁻¹mol⁻¹

Calculation:

101.3 × 20 = 5 × 8.314 × T

2026 = 41.57T

$T = \dfrac{2026}{41.57} = \textbf{49 K}\\\\\text{The Kelvin temperature is }\boxed{\textbf{49 K}}$

6 0

3 years ago

How does a catalyst increase the rate of a chemical reaction?

boyakko [2]

Catalysts decrease the activation energy and the more collisions result in a reaction, so the rate of reaction increases.

4 0

2 years ago

Read 2 more answers

HELP!!!! The freezing of methane is an exothermic change. What best describes the temperature conditions that are likely to make

vazorg [7]

Answer: The correct statement is low temperature only, because entropy decreases during freezing.

Explanation:

The relationship between Gibb's free energy, enthalpy, entropy and temperature is given by the equation:

$\Delta G=\Delta H-T\Delta S$

Where,

$\Delta G$ = change in Gibb's free energy

$\Delta H$ = change in enthalpy

T = temperature

$\Delta S$ = change in entropy

It is given that freezing of methane is taking place, which means that entropy is decreasing and $Delta S$ is becoming negative. It is also given that the reaction is an exothermic reaction, this means that the $\Delta H$ is also negative.

For a reaction to be spontaneous, $\Delta G$ must be negative.

$-ve=-ve-[T(-ve)]\\\\-ve=-ve+T$

From above equations, it is visible that $\Delta G$ will be negative only when the temperature will be low.

Hence, the correct statement is low temperature only, because entropy decreases during freezing.

8 0

3 years ago