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2 years ago

Can anyone help me for these 2 questions

Chemistry

1 answer:

Fynjy0 [20]2 years ago

3 0

1.D 2.A that is pretty hard

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I WILL GIVE BRAINLEST ANSWER THOSE THREE QUESTIONS PLZZZZ HELP ME

miskamm [114]

For changing potential energy; When you drop a book, gravitational potential energy is transformed into kinetic energy. Your car transforms the chemical potential energy stored in gasoline into the kinetic energy of the car's motion.

For Kinetic Energy; kinetic energy can be converted into electrical energy by a generator or into thermal energy by the brakes on a car.

3 0

3 years ago

Determine the mass in grams of 4.69 x 1021 atoms of barium. (The

Virty [35]

Answer:

1.07 g Ba

Explanation:

Hello there!

In this case, according to the definition of the Avogadro's number and the molar mass, it is possible to say that 6.022x10^{23} atoms of barium equal one mole, and at the same time, 1 mole equals 137.327 grams of this element; thus, it is possible to say that 6.022x10^{23} atoms of barium have a mass of 137.327 grams; therefore, it i possible for us to calculate the required mass in grams as shown below:

$4.69x10^{21}atoms*\frac{137.327gBa}{6.022x10^{23} atoms} \\\\=1.07gBa$

Best regards!

5 0

2 years ago

Which of these is a covalent compound?

OLga [1]

A. NO2 because the elements are both nonmetals.

4 0

2 years ago

Read 2 more answers

Balance the following chemical equation by providing the correct coefficients HBr+Ba(OH)2 ———— BaBr2+H2O

stealth61 [152]

Answer:

2HBr + Ba(OH)2 = BaBr2 + 2H20

4 0

2 years ago

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n =

Lady bird [3.3K]

Answer:

$4.86\times10^{-7}\ \text{m}$

Explanation:

R = Rydberg constant = $1.09677583\times 10^7\ \text{m}^{-1}$

$n_1$ = Principal quantum number of an energy level = 2

$n_2$ = Principal quantum number of an energy level for the atomic electron transition = 4

Wavelength is given by the Rydberg formula

$\lambda^{-1}=R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)\\\Rightarrow \lambda^{-1}=1.09677583\times 10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)\\\Rightarrow \lambda=\left(1.09677583\times 10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)\right)^{-1}\\\Rightarrow \lambda=4.86\times10^{-7}\ \text{m}$

The wavelength of the light emitted is $4.86\times10^{-7}\ \text{m}$ .

6 0

2 years ago