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2 years ago

Explain why the Amazon River does not create deltas?

Chemistry

1 answer:

jeka57 [31]2 years ago

4 0

Answer:

The ocean currents are too strong by the Amazon River to form deltas.

Explanation:

The Atlantic has sufficient wave and tidal energy to carry most of the Amazon's sediments out to sea, thus the Amazon does not form a true delta. The great deltas of the world are all in relatively protected bodies of water, while the Amazon empties directly into the turbulent Atlantic.

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Menciona 3 elementos que puedan formar alótropos distintos al carbono

GalinKa [24]

Answer:

tierra agua fuego

Explanation:

espero que esto ayude

8 0

2 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

What is molarity and how is it calculated? If possible, please explain it to me detailed. Thank you :)

Lyrx [107]

Molarity is a concentration unit, defined to be the number of moles of solute divided by the number of liters of solution.

4 0

3 years ago

Carbordum is silicon carbide SiC a very hard material used as an abrasive on sand paper and in other applications. it is prepare

AleksandrR [38]

Answer:

4.5 kilograms of silicon dioxide is required to produce 3.00 kg of SiC.

Explanation:

The balanced equation for the reaction between silicon dioxide and carbon at high temperature is given as:

$SiO_2+3C\rightarrow SiC+2CO$

1 mole silicon dioxide reacts with 3 moles of carbon to give 1 moles of silicon carbide and 2 moles of carbon monoxide.

Mass of SiC = 3.00kg = 3000.00 g

1 kg = 1000 g

Molecular mass of SiC = 40 g/mol

Moles of SiC = $\frac{3000.00 g}{40 g/mol}= 75 mol$

According to reaction, 1 mole of SiC is produced from 1 mole of silicon dioxide.

Then 75 moles of SiC will be produce from:

$\frac{1}{1}\times 75 mol=75 mol$ of silicon dioxide.

mass of 75 moles of silicon dioxde:

$75 mol\times 60 g/mol=4500 g=4.5 kg$

4.5 kilograms of silicon dioxide is required to produce 3.00 kg of SiC.

3 0

3 years ago

How many Atoms of Oxygen are in water H20? explain how u know

blagie [28]

There are ALOT because they would always come in and out and they will burst which creates more so techneclly there are infinate

3 0

3 years ago

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