Sunspots sit on the sun’s

How many grams of sulfur are required in the preparation of 35.7 moles of sulfur dioxide Sg + O2 SO2

lidiya [134]

S+O_2-->SO_2

Its balanced already

So

1 mol SO_2 require 1 mole sulphur
35.7moles require 35.7mol sulphur

Mass of S

35.7(32)
1142.4g

6 0

2 years ago

How many dimes could you trade for 360 pesos? $1 = 1500 pesos.

sveta [45]

Answer:

The correct option is (d).

Explanation:

It is given that,

1$ = 1500 pesos

We need to convert 360 pesos into dimes

We can convert 360 pesos to dollars as follows:

$360\ \text{pesos}=\$\dfrac{1}{1500}\times 360\\\\=$0.24$

360 pesos is equal to $0.24

Also, 1 dollar = 10 dimes

We can covert 0.24 dollar to dimes as follows :

0.24 dollar = 10 × 0.24 dimes

0.24 dollar = 2.4 dimes

or

360 pesos = 2.4 dimes

7 0

3 years ago

In acid solution, water can add to the double bond of 2-butenedioic acid to form 2-hydroxysuccinic acid.

satela [25.4K]

So, what is the question?

6 0

3 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

Equal amounts of methylene blue (319.85 g/mol) and potassium permanganate (158.034 g/mol) dyes were added to identical agar plat

tamaranim1 [39]

Answer:

Potassium permanganate.

Explanation:

Both substances are dyes, but the methylene blue has a bigger molecular mass (319.85 g/mol), that means that the particles are bigger in comparison with the potassium permanganate that has a molecular mass of 158.034 g/mol.

Since the molar mass is the half in the case of potassium permanganate, it can be considered that the particle size is the half in size. In the agar, a smaller particle will present less resistance to flow, that means that it going to move faster.

4 0

3 years ago