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3 years ago

According to the law of conservation of mass, which of the pictured statements is true?

Chemistry

1 answer:

kirill115 [55]3 years ago

4 0

Answer:

both the mass number and atomic nbets have to add up to he equal to each other on either side of the arrow. thanks

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An ethylene glycol solution contains 16.2 g of ethylene glycol (c2h6o2) in 85.4 ml of water. calculate the boiling point of the

Andre45 [30]

<span>Answer: (16.2 g C2H6O2) / (62.0678 g C2H6O2/mol) / (0.0982 kg) = 3.9704 mol/kg = 3.9704 m a.) (3.9704 m) x (1.86 Â°C/m) = 7.38 Â°C change 0.00Â°C - 7.38 Â°C = - 7.38 Â°C b.) (3.9704 m) x (0.512 Â°C/m) = 2.03 Â°C change 100.00Â°C + 2.03 Â°C = 102.03 Â°C</span>

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4 years ago

After one half life ___% of the atoms will change to a stable

Nezavi [6.7K]

Answer:

the answer is 30% if the atoms will change to a stable

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3 years ago

Study the graph about seismic waves.

Salsk061 [2.6K]

Answer:

siezmic waves increase

Explanation:

5 0

3 years ago

Read 2 more answers

45 The mass of an unidentified metal sphere is 133 grams. Students determine the

Lisa [10]

Answer:

D Cobalt

Explanation:

The volume of the sphere is 40 -25 = 15 cm^3

Density = mass/volume = 133 gm / 15 cm^3 = 8.87 gm/cm^3

which corresponds to Cobalt from the chart

7 0

2 years ago

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands

Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

$E=\frac{R_y}{n^2}$

$R_y=2.18\times 10^{-18} J$ = Rydberg energy

n = principal quantum number of the orbital

Energy of 11th orbit = $E_{11}$

$E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J$

Energy of 10th orbit = $E_{10}$

$E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J$

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

$E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J$

$=E'=0.38\times 10^{-20} J$

$\lambda =\frac{hc}{E'}$ (Planck's' equation)

$\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}$

$\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m$

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

3 0

3 years ago