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3 years ago

In which type of climate would the soil most likely be well developed and have lots of organic matter?

1 answer:

3 0

I suppose it would be forest because in order to have organic matter the soil needs to be rich and fertile,therefore it is forest.

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What charges must the particles have to create the electric field shown?

Law Incorporation [45]

Answer: B Explanation: The reason is u can’t have a stable current without a positive and negative.

6 0

2 years ago

Acetylene (C2H2) gas is often used in welding torches because of the very high heat produced when it reacts with oxygen (O2) gas

babymother [125]

Answer:

0.05 mol

Explanation:

The balanced equation for the reaction that takes place is:

2C₂H₂ (g) + 5O₂ (g) → 4CO₂ (g) + 2H₂O (g)

Now we convert 0.10 moles of carbon dioxide (CO₂) into moles of acetylene (C₂H₂), using the stoichiometric coefficients of the balanced reaction:

0.10 mol CO₂ * $\frac{2molC_2H_2}{4molCO_2}$ = 0.05 mol C₂H₂

7 0

3 years ago

What is the difference between ionic and covalent bonds? Give an example of a each type

Norma-Jean [14]

Answer:

Ionic bonds are opposite, whereas Covalent bonds are opposite.

Explanation:

Two opposite ions attract each other and form the ionic bond. Covalent bonding is a form of chemical bonding between two non metallic atoms which is characterized by the sharing of pairs of electrons between atoms and othercovalent bonds.

Examples:

Sodium Chloride and Sodium Fluoride are examples of Ionic bonds.

and

Water and Carbon Dioxide are examples of Covalent bonds. 

4 0

4 years ago

Ammonia, chlorine and hydrogen chloride cannot be collected by displacement of water. Why?

Archy [21]

Answer:

Yes

Explanation:

Because they are lighter than water

4 0

3 years ago

If our eyes could see a slightly wider region of the electromagnetic spectrum, we would see a fifth line in the Balmer series em

erica [24]

Answer: Wavelength associated with the fifth line is 397 nm

Explanation:

$E=\frac{hc}{\lambda}$

$\lambda$ = Wavelength of radiation

E= energy

For fifth line in the H atom spectrum in the balmer series will be from n= 2 to n=7.

Using Rydberg's Equation:

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

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $10973731.6m^{-1}$

$n_f$ = Higher energy level = 7

$n_i$ = Lower energy level = 2 (Balmer series)

Putting the values, in above equation, we get

$\frac{1}{\lambda}=10973731.6\times \left(\frac{1}{2^2}-\frac{1}{7^2} \right )\times 1^2$

$\frac{1}{\lambda}=2.52\times 10^{6}m$

$\lambda}=3.97\times 10^{-7}m=397 nm$ $1nm=10^{-9}m$

Thus wavelength λ associated with the fifth line is 397 nm

7 0

4 years ago