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3 years ago

Carbonic acid dissolves limestone and other rocks. This is an example of _____. chemical errosion

1 answer:

8 0

Carbonic acid dissolves limestone and other rocks. This is an example of chemical erosion. An example is in the caves. Caves are formed where rainwater as it falls through the atmosphere absorbs carbon dioxide. The carbon dioxide makes the rain acidic to react it with the limestone bedrock. The rainwater is absorbed by the soil into the ground. Then as it enters through the soil, the rainwater will absorb more carbon dioxide that is produced by the decomposers. The carbon dioxide with water reacts to form carbonic acid. The carbonic acid will react to limestone and dissolves it slowly. As the space become larger, water can enter into it.

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the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal

Sunny_sXe [5.5K]

Rydberg formula is given by:

$\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )$

where, $R_{H}$ = Rydberg constant = $1.0973731568508 \times 10^{7} per metre$

$\lambda$ = wavelength

$n_{1}$ and $n_{2}$ are the level of transitions.

Now, for $n_{1}$ = 2 and $n_{2}$ = 6

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )$

= $1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0278 )$

= $1.0973731568508 \times 10^{7} \times 0.23$

= $0.2523958\times 10^{7}$

$\lambda = \frac{1}{0.2523958\times 10^{7}}$

= $3.9620\times 10^{-7} m$

= $396.20\times 10^{-9} m$

= $396.20 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 5

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.04 )$

= $1.0973731568508 \times 10^{7} \times (0.21 )$

= $0.230 \times 10^{7}$

$\lambda= \frac{1}{0.230 \times 10^{7}}$

= $4.3478 \times 10^{-7} m$

= $434.78\times 10^{-9} m$

= $434.78 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 4

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0625 )$

= $1.0973731568508 \times 10^{7} \times (0.1875 )$

= $0.20575 \times 10^{7}$

$\lambda= \frac{1}{0.20575 \times 10^{7}}$

= $4.8602 \times 10^{-7} m$

= $486.02 \times 10^{-9} m$

= $486.02 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 3

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.12 )$

= $1.0973731568508 \times 10^{7} \times (0.13 )$

= $0.1426585\times 10^{7}$

$\lambda= \frac{1}{0.1426585\times 10^{7}}$

= $7.0097 \times 10^{-7} m$

= $700.97 \times 10^{-9} m$

= $700.97 nm$

5 0

2 years ago

Read 2 more answers

18 g of argon occupy 750 ml at a particular temperature and pressure. How many grams of methane would occupy the same volume at

frozen [14]

Answer:

7.21 grams is the mass of methane

Explanation:

We may use the Ideal Gases Equation to solve this:

P. V = n. R. T

Let's determine the moles of Ar

18 g . 1 mol/ 39.9 g = 0.451 mol

In both situations, volume, temperature and pressure are the same so the moles of methane will also be the same as Argon's.

Let's convert the moles to mass of CH4.

0.451 mol . 16g/1mol = 7.21 grams

5 0

3 years ago

How many grams of H2O will be formed when 36.8 g H2 is mixed with 40.2 g O2 and allowed to completely react to form water

Artemon [7]

Answer:

45.225 grams of H₂O will be formed when 36.8 g H₂ is mixed with 40.2 g O₂

Explanation:

The balanced reaction is:

2 H₂ + O₂ → 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

H₂: 2 moles
O₂: 1 mole
H₂O: 2 moles

Being the molar mass of the compounds:

H₂: 2 g/mole
O₂: 32 g/mole
H₂O: 18 g/mole

then, by reaction stoichiometry, the following amounts of reactant and product mass participate:

H₂: 2 moles* 2 g/mole= 4 g
O₂: 1 mole* 32 g/mole= 32 g
H₂O: 2 moles* 18 g/mole= 36 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction and a simple rule of three as follows: if by stoichiometry 4 g of H₂ react with 32 g of O₂, 36.8 g of H₂ with how much mass of O₂ will it react?

$mass of O_{2} =\frac{36.8 grams of H_{2}*32 grams of O_{2} }{4 grams of H_{2}}$

mass of O₂=294.4 grams

But 294.4 grams of O₂ are not available, 40.2 grams are available. Since you have less mass than you need to react with 36.8 grams of H₂, oxygen O₂ will be the limiting reagent.

Then you can apply the following rule of three: if by stoichiometry 32 grams of O₂ form 36 grams of H₂O, 40.2 grams of O₂ how much mass of H₂O will it form?

$mass of H_{2}O=\frac{40.2 grams of O_{2} *36 grams of H_{2}O }{32 grams of O_{2} }$

mass of H₂O= 45.225 grams

45.225 grams of H₂O will be formed when 36.8 g H₂ is mixed with 40.2 g O₂

8 0

2 years ago

a. silicon b. carbon c. beryllium d.chromium

zepelin [54]

The answer is silicon because it’s atomic number is 14

3 0

3 years ago

Read 2 more answers

What is the number of moles of solute in 250 mL of a 0.4 M solution?

mafiozo [28]

Answer:

0,1 mol

Explanation:

We know that the formula of concentration is C= moles of solute/ volume

0,4 M= moles of solute/ 250 mL

Convert mL to L 250 mL =0,25 L

0,4 M x 0,25 L= moles of solute

0,1 moles= moles of solute

3 0

3 years ago