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3 years ago

5. Which is an example of inertia?

Chemistry

1 answer:

Hitman42 [59]3 years ago

6 0

Answer:

Explanation:

D. When the truck driver slammed on the brakes, all the boxes in the back

of the truck slid forward.

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An atom has 8 protons, 8 neutrons and 9 electrons. is thus a cation, anion, or neutral atom?

Black_prince [1.1K]

Answer:

It is an Anion because there are more electrons than protons making it negative

7 0

3 years ago

What is the volume of the sample at stp ?

Natasha2012 [34]

At stp the volume is 22.4 L .

hope this helps!

3 0

3 years ago

What volume of 0.305 m agno3 is required to react exactly with 155.0 ml of 0.274 m na2so4 solution? hint: you will want to write

LuckyWell [14K]

The balanced chemical equation for reaction of $AgNO_{3}$ and $Na_{2}SO_{4}$ is as follows:

$2 AgNO_{3}+Na_{2}SO_{4}\rightarrow 2NaNO_{3}+Ag_{2}SO_{4}$

From the balanced chemical equation, 2 mol of $AgNO_{3}$ reacts with 1 mol of $NaNO_{3}$ .

First calculating number of moles of $NaNO_{3}$ as follows:

$M=\frac{n}{V}$

On rearranging,

$n=M\times V$

Here, M is molarity and V is volume. The molarity of $NaNO_{3}$ is given 0.274 M or mol/L and volume 155 mL, putting the values,

$n=0.274 mol/L\times 155\times 10^{-3}mL=0.04247 mol$

Since, 1 mol of $NaNO_{3}$ reacts with 2 mol of $AgNO_{3}$ thus, number of moles of $AgNO_{3}$ will be $2\times 0.04247 mol=0.08494 mol$ .

Now, molarity of $AgNO_{3}$ is given 0.305 M or mol/L thus, volume can be calculated as follows:

$V=\frac{n}{M}=\frac{0.08494 mol}{0.305 mol/L}=0.2785 L=278.5 mL$

Therefore, volume of $AgNO_{3}$ is 278.5 mL.

4 0

3 years ago

During a reaction in an aqueous solution, the concentration of bactants

STALIN [3.7K]

Answer:

my define it will be turst me is c

5 0

3 years ago

Consider a transition of the electron in the hydrogen atom from n=3 to n=7.

kow [346]

Answer:

For a: The wavelength of light is $1.005\times 10^{-6}m$

For b: The light is getting absorbed

Explanation:

For a:

To calculate the wavelength of light, we use Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.097\times 10^7m^{-1}$

$n_f$ = Higher energy level = 7

$n_i$ = Lower energy level = 3

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{3^2}-\frac{1}{7^2} \right )\\\\\lambda =1.005\times 10^{-6}m$

Hence, the wavelength of light is $1.005\times 10^{-6}m$

For b:

There are two ways in which electrons can transition between energy levels:

Absorption spectra: This type of spectra is seen when an electron jumps from lower energy level to higher energy level. In this process, energy is absorbed.
Emission spectra: This type of spectra is seen when an electron jumps from higher energy level to lower energy level. In this process, energy is released in the form of photons.

As, the electron jumps from lower energy level to higher energy level. The wavelength is getting absorbed.

6 0

3 years ago