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2 years ago

Why aren’t the same natural resources found all over the world?

2 answers:

7 0

Answer: OK so Natural resources are not evenly distributed all over the world. Some places are more endowed than others — for instance, some regions have lots of water (and access to the ocean and seas). Others have lots of minerals and forestlands. Others have metallic rocks, wildlife, fossil fuels, and so on.The distribution of natural resources depends upon many physical factors like land, climate and altitude. The distribution of resources is unequal because these factors differ from place to place on this earth. Hope this helps have a nice night❤️

Explanation:

soldi70 [24.7K]2 years ago

3 0

Answer:

if there was there would be no need for traveling

Explanation:

You might be interested in

How does the size change when an atom forms a cation and when an atom forms an anion?

mario62 [17]

<span>A cation is an atom that loses a valence electron. When a valence electron is released there is one electron less to create a repulsive force. The loss of a repulsive force will allow the atom to pull tighter together. An anion would therefore be larger in size due to increased repulsion of the valence electrons.</span>

8 0

3 years ago

Assuming you have 6.24 x 1014 electrons and the surface area of the pail is 0.2 m2, what is the charge density (C/m2)?

pentagon [3]

Answer:

σ = 4.998 E-4 C/m²

Explanation:

1 Coulomb (C) ≡ 6.241509 E18 electrons (e)

∴ # elect = 6.24 E14 elect

charge (Q):

⇒ Q = (6.24 E14 elect)/( 1 C /6.241509 E18 elect) = 9.998 E-5 C

charge density (σ):

σ = Q/S

∴ surface area (S) = 0.2 m²

⇒ σ = ( 9.998 E-5 C ) / ( 0.2 m²)

⇒ σ = 4.998 E-4 C/m²

4 0

2 years ago

The molarity of a 2 liter aqueous solution that contains 222.2 grams of dissolved calcium chloride ( CaCl2), expressed with two

sveta [45]

Answer:

The answer to your question is 1 M

Explanation:

Data

Molarity = ?

mass of CaCl₂ = 222.2 g

Volume = 2 l

Process

1.- Calculate the molar mass of CaCl₂

CaCl₂ = 40 + (35.5 x 2) = 40 + 71 = 111 g

2.- Calculate the moles of CaCl₂

111g of CaCl₂ ---------------- 1 mol

222.2 f of CaCl₂ ---------------- x

x = (222.2 x 1) / 111

x = 222.2 / 111

x = 2 moles

3.- Calculate the Molarity

Molarity = moles / Volume

-Substitution

Molarity = 2/2

-Result

Molarity = 1

3 0

3 years ago

A habit is: <br>pls help me ty

kherson [118]

What class is this? (Subject)

4 0

2 years ago

Read 2 more answers

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