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2 years ago

How many moles of h2 gas would be contained in 4.00 of the gas at 202.6KPa and 127c?

Physics

1 answer:

JulsSmile [24]2 years ago

4 0

Answer : The volume of one mole of gas is dependent of temperature and pressure.

Explanation:

At 0oCand1atm one mole of an ideal gas would be ≈22.4L, which is the same as 22.4dm3

If you double the pressure, the same amount (mole) of gas would take up only half the space.

If the temperature changes (and pressure stays constant), the gas will expand or contract by about 1%/3o.

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An uncrewed mission to the nearest star, Proxima Centauri, is launched from the Earth's surface as a projectile with an initial

Anna [14]

Answer:

42.96 km/s

Explanation:

From the conservation of Energy

$(PE+KE)_i=(PE+KE)_f\\\Rightarrow -\frac{GmM}{R}+\frac{1}{2}mv_i^2=0+\frac{1}{2}mv_f^2$

Mass gets cancelled

$-\frac{GM}{R}+\frac{1}{2}v_i^2=0+\frac{1}{2}v_f^2\\\Rightarrow -2\frac{GM}{R}+v_i^2=v_f^2\\\Rightarrow -v_e^2+v_i^2=v_f^2\\\Rightarrow v_f=\sqrt{v_i^2-v_e^2}$

$v_e=\sqrt{\frac{2Gm}{R}}$ = Escape velocity of Earth = 11.2 km/s

$v_i$ = Velocity of projectile = 44.4 km/s

$v_f=\sqrt{44.4^2-11.2^2}\\\Rightarrow v_f=42.96\ km/s$

The velocity of the spacecraft when it is more than halfway to the star is 42.96 km/s

6 0

2 years ago

The ratio of carbon-14 to nitrogen-14 is an artifact is 1:3. Given that half-life of carbon-14 is 5730years, how old is the arti

dusya [7]

Answer:

9155 years old

Explanation:

We use the following expression for the decay of a substance:

$N = N_0\,\,e^{-k*t}$

So we first estimate the value of k knowing that the half-life of the C14 is 5730 years:

$N = N_0\,\,e^{-k*t}\\N_0/2=N_0\,\,e^{-k*5730}\\1/2 = e^{-k*5730}\\ln(1/2)=-k*5730\\k= 0.00012$

so, now we can estimate the age of the artifact by solving for"t" in the equation:

$1/3=e^{-0.00012*t}\\ln(1/3)= -0.00012*t\\t=9155. 102$

which we can round to 9155 years old.

5 0

3 years ago

A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w

spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

directly proportional to its length i.e $R\propto l$
inversely proportional to its cross section area i.e $R\propto \frac{1}{A}$

Therefore

$R=\rho\frac{l}{A}$

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

$A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2$

$R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }$

$\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }$ ......(1)

Again for tungsten:

$R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }$

$\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }$ ........(2)

Given that $R_1=R_2$ and $l_1=l_2$

Dividing the equation (1) and (2)

$\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}$

$\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}$ [since $R_1=R_2$ and $l_1=l_2$ ]

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}$

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}$

$\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}$

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

8 0

3 years ago

What do folds on a strip of paper mean

Vlad [161]

There are creases. Or just origami

7 0

2 years ago

An object travels with velocity v = 4.0 meters/second and it makes an angle of 60.0° with the positive direction of the y-axis.

zhenek [66]

Answer:

2 m/s and -2 m/s

Explanation:

The object travels with an angle of

60.0°

with the positive direction of the y-axis: this means that it lies either in the 1st quadrant (positive x) or in the 2nd quadrant (negative x).

If it lies in the 1st quadrant, the value of vx (component of v along x direction) is:

$v_x = v cos \theta = (4.0 m/s) cos 60.0^{\circ}=2 m/s$

If it lies in the 2nd quadrant, the value of vx (component of v along x direction) is:

$v_x = -v cos \theta = -(4.0 m/s) cos 60.0^{\circ}=-2 m/s$

5 0

3 years ago