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2 years ago

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According to the Law of the Conservation of Matter, if you dissolve 25 grams of sugar into 150 grams of water, the mixture shoul

d weigh how much?

1 answer:

levacccp [35]2 years ago

7 0

19grams becauce if u

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Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun

iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

$E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})$ (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

$E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV$

B. The energy of the emitted photon is given by the following formula:

$E=h\frac{c}{\lambda}$ (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

$2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J$

Next, you use the equation (2) and solve for λ:

$\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm$

C. The radius of the orbit is given by:

$r_n=n^2a_o$ (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

$r_5=5^2(2.380)=59.5$

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0

2 years ago

Convection currents in air and water occur because _

Diano4ka-milaya [45]

warm fluids are less dense than cold fluids

7 0

2 years ago

In 2h2so4, the number of hydrogen atoms is 1 2 4 8

pishuonlain [190]

C.) In this, number of Hydrogen atoms is 4

7 0

2 years ago

The equation for momentum is p=mv. Which letter represents the momentum? Question 2 options: p m v

vovikov84 [41]

P stands for momentum

8 0

3 years ago

You need to produce a set of cylindrical copper wire 3.5 m long that will have a

vazorg [7]

Solution :

We know, resistance is given by :

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

$A = \dfrac{\rho l }{R}\\\\A = \dfrac{1.72\times 10^{-8} \times 3.5 }{0.125}\\\\A = 4.816 \times 10^{-7} \ m^2$

Now, we know mass of wire is given by :

$Mass = Density \times Volume\\\\\M = 8.9 \times 10^3 \times 4.816 \times 10^{-7} \times 3.5 kg\\\\M = 0.01500\ kg\\\\M = 15.00\ gram$

Hence, this is the required solution.

8 0

2 years ago