Density is ———- Through the world’s ocean

Electrons moving back and forth through a circuit would be

yaroslaw [1]

The particles that carry charge through wires in a circuit are mobile electrons. The electric field direction within a circuit is by definition the direction that positive test charges are pushed. Thus, these negatively charged electrons move in the direction opposite the electric field.

5 0

4 years ago

A gas mixture contains 320mg methane, 175 mg argon, 225 mg nitrogen (N2). The partial pressure of argon at 300K is 12.52 kPa. Wh

svet-max [94.6K]

Answer: The volume and total pressure of the mixture is 0.876 L and 89.36 kPa respectively.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For methane:

Given mass of methane = 320 mg = 0.3 g (Conversion factor: 1 g = 1000 mg)

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane}=\frac{0.3g}{16g/mol}=0.019mol$

For argon:

Given mass of argon = 175 mg = 0.175 g

Molar mass of argon = 40 g/mol

Putting values in equation 1, we get:

$\text{Moles of argon}=\frac{0.175g}{40g/mol}=0.0044mol$

For nitrogen:

Given mass of nitrogen = 225 mg = 0.225 g

Molar mass of nitrogen = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen}=\frac{0.225g}{28g/mol}=0.0080mol$

To calculate the volume of the mixture, we use the equation:

PV = nRT ......(2)

We are given:

Partial pressure of argon = 12.52 kPa

Temperature = 300 K

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

n = number of moles of argon = 0.0044 moles

Putting values in equation 2, we get:

$12.52kPa\times V=0.0044mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times 300K\\\\V=\frac{0.0044\times 8.31\times 300}{12.52}=0.876L$

Now, calculating the total pressure of the mixture by using equation 2:

Total number of moles = [0.019 + 0.0044 + 0.0080] mol = 0.0314 mol

V= volume of the mixture = 0.876 L

Putting values in equation 2, we get:

$P\times 0.876L=0.0314mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times 300K\\\\P=\frac{0.0341\times 8.31\times 300}{0.876}=89.36kPa$

Hence, the volume and total pressure of the mixture is 0.876 L and 89.36 kPa respectively.

5 0

3 years ago

What is the velocity of position A and B? answer please

BaLLatris [955]

Answer:

A = 15 m/s , B = 18.75 m/s

Explanation:

from the velocity is equal to zero ( at rest ) , you see that velocity is increasing by 3.75 m/s for each second.

I hope that it's a correct answer.

5 0

3 years ago

What happens to the atomic number of an atom when the number of neutrons in the nucleus of that atom increases? a It decreases b

kupik [55]

Answer:

It remains the same

Explanation:

It remains the same. This is because the number of protons doesn't change and the number of protons determines the atomic number.

8 0

2 years ago

A laser used for many applications of hard surface dental work emits 2780-nm wavelength pulses of variable energy (0-300 mJ) abo

Bess [88]

Answer:

a

$n = 1.119 *10^{18} \ photons$

b

$P = 1.6 \ W$

Explanation:

From the question we are told that

The wavelength is $\lambda = 2780 nm = 2780 *10^{-9} \ m$

The energy is $E = 80 mJ = 80 *10^{-3} \ J$

This energy is mathematically represented as

$E = \frac{n * h * c }{\lambda }$

Where c is the speed of light with a value $c = 3.0 *10^{8} \ m/s$

h is the Planck's constant with the value $h = 6.626 *10^{-34} \ J \cdot s$

n is the number of pulses

So

$n = \frac{E * \lambda }{h * c }$

substituting values

$n = \frac{80 *10^{-3} * 2780 *10^{-9}}{6.626 *10^{-34} * 3.0 *10^{8} }$

$n = 1.119 *10^{18} \ photons$

Given that the pulses where emitted 20 times in one second then the period of the pulse is

$T = \frac{1}{20}$

$T = 0.05 \ s$

Hence the average power of photons in one 80-mJ pulse during 1 s is mathematically represented as

$P = \frac{E}{T}$

substituting values

$P = \frac{ 80 *10^{-3}}{0.05}$

$P = 1.6 \ W$

6 0

4 years ago