What is the number of elements in 2Ca(OH)2

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11111nata11111 [884]

The correct answer is B. Homeostasis

5 0

2 years ago

I need it please !!!

anyanavicka [17]

Answer:

im pretty sure its 2...

Explanation:

if its wrong im sorry

if its right brainliest pls?

6 0

2 years ago

When nitrogen dioxide (NO2) from car exhaust combines with water in the air, it forms nitric acid (HNO3), which causes acid rain

yKpoI14uk [10]

Answer:

For a: The number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

For b: The mass of nitric acid formed is 54.81 grams

For c: The mass of nitric acid formed is 206 grams

Explanation:

The given chemical reaction follows:

$3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)$

For a:

By Stoichiometry of the reaction:

1 mole of water reacts with 3 moles of nitrogen dioxide

So, 0.250 moles of water will react with $\frac{3}{1}\times 0.250=0.75mol$ of nitrogen dioxide

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.75 moles of nitrogen dioxide will contain $0.75\times 6.022\times 10^{23}=4.52\times 10^{23}$ number of molecules

Hence, the number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

For b:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of nitrogen dioxide = 60.0 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{60.0g}{46g/mol}=1.304mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide produces 2 mole of nitric acid

So, 1.304 moles of nitrogen dioxide will produce = $\frac{2}{3}\times 1.304=0.870$ moles of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 0.870 moles

Putting values in equation 1, we get:

$0.870mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(0.870mol\times 63g/mol)=54.81g$

Hence, the mass of nitric acid formed is 54.81 grams

For c:

For nitrogen dioxide:

Given mass of nitrogen dioxide = 225 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{225g}{46g/mol}=4.90mol$

For water:

Given mass of water = 55.2 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{55.2g}{18g/mol}=3.06mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide reacts with 1 mole of water

So, 4.90 moles of nitrogen dioxide will react with = $\frac{1}{3}\times 4.90=1.63mol$ of water

As, given amount of water is more than the required amount. So, it is considered as an excess reagent.

Thus, nitrogen dioxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 mole of nitrogen dioxide produces 2 moles of nitric acid

So, 4.90 moles of nitrogen dioxide will produce $\frac{2}{3}\times 4.90=3.27mol$ of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 3.27 moles

Putting values in equation 1, we get:

$3.27mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(3.27mol\times 63g/mol)=206g$

Hence, the mass of nitric acid formed is 206 grams

7 0

3 years ago

C -14 and N-14 both have same mass number yet they are different elements. Explain

KiRa [710]

Explanation:

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7 0

3 years ago

Read 2 more answers

Calculate the energy for the transition of an electron from the n = 5 level to the n = 6 level of a hydrogen atom. E = Joules Is

kramer

Answer:

For an electron to move from a lower energy level to a higher energy , that electron needs to absorb energy sufficient enough to excite it to make the transition. Hence it is an absorption process. The required energy of transition E = 2.665 x 10⁻²⁰J

Explanation:

Using the Rydberg's equation we can calculate the wavelength of the photon of energy transition as follows:

1/λ = R . (1/nf² - 1/ni²)

where

λ is the required wavelength of the photon needed to be absorbed to excite the electron to transit from level 5 to 6.

(Note that for the electron to transit to from energy level 5 to 6, the photon would have to fall from level 6 to 5 in order to emit the required energy to excite the electron)

R is the Rydberg's constant 1.097 x 10⁷ m⁻¹

nf is the final level of the photon

ni is the initial level of the photon

1/λ = 1.097 x 10⁷ m⁻¹ (1/5² - 1/6²)

1/λ = 1.3407 x 10⁵ m⁻¹

λ = 7.458 x 10⁻⁶ m

This implies that that is the wavelength of the photon required to excite the electron to transit from energy level 5 to 6. Using the equation below, we can calculate the energy of transition as

E = h.c/λ

where

E is the required energy of transition

h is the Planck's constant (6.626 x 10⁻³⁴ Js)

c is the speed of light (3 x 10⁸ms⁻¹)

λ is the wavelength calculated above

E = 6.626 x 10⁻³⁴ Js x 3 x 10⁸ms⁻¹/ 7.458 x 10⁻⁶ m

E = 2.665 x 10⁻²⁰J

3 0

3 years ago