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4 years ago

The gram formula mass of a compound is 48 grams. The mass of 1.0 moles of this compound is

2 answers:

8 0

Gram formula mass means the mass of one moles of the compound
so The mass of 1.0 moles of this compound is 48g
so i conclude option 3 is correct
hope it helps

Vesna [10]4 years ago

3 0

Answer:

The mass of 1.0 moles of this compound is (3) 48g.

Explanation:

The gram formula mass is defined as the amount of mass in 1.0 moles of a compound.

For example :

For the water, we can calculate its gram formula mass using the molecular formula of the water : $H_{2}O$

To calculate the gram formula mass we need the molecular mass of H (hydrogen) and O (oxygen).

μ(H) = $1\frac{g}{mole}$

μ(O) = $16\frac{g}{mole}$ ⇒ The gram formula mass of water is:

2 [ μ(H) ] + μ(O) = $2(1\frac{g}{mole})+16\frac{g}{mole}=18\frac{g}{mole}$

So 1 mole of water contains 18 grams of water.

In the exercise, the gram formula mass of a compound is 48 grams. Therefore, 1 mole of the compound contains 48 g of compound.

Finally, the correct option is (3) 48g

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A 12,500 N alien UFO is hovering about the surface of Earth. At time , its position can be given as () = ((0.24 m/s^3)^3 + 25 m)

stepladder [879]

a) $F=(3675i-4543k)N$

b) 5843 N

Explanation:

The position of the UFO at time t is given by the vector:

$r(t)=(0.24t^3+25)i+(4.2t)j+(-0.43t^3+0.8t^2)k$

Therefore it has 3 components:

$r_x=0.24t^3+25\\r_y=4.2t\\r_z=-0.43t^3+0.8t^2$

We start by finding the velocity of the UFO, which is given by the derivative of the position:

$v_x=r'_x=\frac{d}{dt}(0.24t^3+25)=3\cdot 0.24t^2=0.72t^2\\v_y=r'_y=\frac{d}{dt}(4.2t)=4.2\\v_x=r'_z=\frac{d}{dt}(-0.43t^3+0.8t^2)=-1.29t^2+1.6t$

And then, by differentiating again, we find the acceleration:

$a_x=v'_x=\frac{d}{dt}(0.72t^2)=1.44t\\a_y=v'_y=\frac{d}{dt}(4.2)=0\\a_z=v'_z=\frac{d}{dt}(-1.29t^2+1.6t)=-2.58t+1.6$

The weight of the UFO is W = 12,500 N, so its mass is:

$m=\frac{W}{g}=\frac{12500}{9.8}=1276 kg$

Therefore, the components of the force on the UFO are given by Newton's second law:

$F=ma$

So, Substituting t = 2 s, we find:

$F_x=ma_x=(1276)(1.44t)=(1276)(1.44)(2)=3675 N\\F_y=ma_y=0\\F_z=ma_z=(1276)(-2.58t+1.6)=(1276)(-2.58(2)+1.6)=-4543 N$

So the net force on the UFO at t = 2 s is

$F=(3675i-4543k)N$

The magnitude of a 3-dimensional vector is given by

$|v|=\sqrt{v_x^2+v_y^2+v_z^2}$

where

$v_x,v_y,v_z$ are the three components of the vector

In this problem, the three components of the net force are:

$F_x=3675 N\\F_y=0\\F_z=-4543 N$

Therefore, substituting into the equation, we find the magnitude of the net force:

$|F|=\sqrt{3675^2+0^2+(-4543)^2}=5843 N$

7 0

3 years ago

If a professional football player and a toddler were both running at you, which would be easier to stop? Why? Use the terms forc

igor_vitrenko [27]

Answer:

The todler just common sense

sorry if im wrong

Explanation:

4 0

3 years ago

Read 2 more answers

34000×0.4 ---------------- 0.02×200 In scientific notation

VashaNatasha [74]

Answer:

the simplified expression is written as 3.4 x 10³

Explanation:

Given expression;

$\frac{34000\times 0.4}{0.02 \times 200}$

in scientific notation, the expression is simplified as;

$\frac{34000\times 0.4}{0.02 \times 200} = \frac{13600}{4} = 3400 = 3.4 \times 10^3$

Therefore, in scientific notation, the simplified expression is written as 3.4 x 10³

4 0

3 years ago

What happens when an electron moving from the 3rd energy level to the 1st energy level?

Sholpan [36]

Answer:

A photon of wavelength 103 nm is released

Explanation:

When an electron in an atom jumps from a higher energy level to a lower energy level, it releases a photon whose energy is equal to the difference in energy between the two levels.

For example, if we are talking about a hydrogen atom, the energy of the levels are:

$E_1 = -13.6 eV\\E_2 = -3.4 eV\\E_3 = -1.5 eV$

So, the energy of the photon released when the electron jumps from the level n=3 to n=1 is

$\Delta E = E_3 - E_1 = -1.5 -(-13.6)=12.1 eV$

In Joules,

$\Delta E =12.1\cdot 1.6\cdot 10^{-19} = 1.94\cdot 10^{-18} J$

We can also find the wavelength of this photon, using the equation:

$\Delta E = \frac{hc}{\lambda}\rightarrow \lambda=\frac{hc}{\Delta E}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{1.94\cdot 10^{-18}}=1.03\cdot 10^{-7} m = 103 nm$