All categories

3 years ago

Which elements are found in a high portion of earths crust

2 answers:

8 0

The elements found in a high portion of earths crust are Oxygen, Aluminum, Iron, Calcium, Potassium, Sodium and magnesium

Sonbull [250]3 years ago

3 0

The elements that are found in a high proportion in Earth's crust are oxygen, aluminium, iron, calcium, potassium, sodium, andmagnesium. this is because oxygen makes 46.6% while Silicon makes 27.7% of the Earth crust.

hope it helps

You might be interested in

IDENTIFY THE CHEMICAL REACTIONS( CLICK ALL THAT APPLY)

AlladinOne [14]

Answer:

B. single replacement

Explanation:

5 0

4 years ago

Read 2 more answers

Hydrogen chloride gas reacts with solid silicon

oee [108]

Liquid disilicon hexachloride reacts with water to form solid silicon dioxide, hydrogen chloride gas, and hydrogen gas.

7 0

3 years ago

How many moles of sodium are present in 17 g of Na?

11Alexandr11 [23.1K]

Answer:

1. 0.74mol

2. 0.42mol

3. 2.125mol

4. 0.301mol

5. 4.52 × 10^23 particles

Explanation:

Number of moles (n) in a substance can be found using the formula:

mole (n) = mass/molar mass

Using this formula, the following moles are calculated:

1. Molar of Na = 23g/mol

mole = 17/23

mole = 0.74mol

2. Molar mass of Na2SO4 = 23(2) + 32 + 16(4)

= 46 + 32 + 64

= 142g/mol

Mole = 60/142

mole = 0.42mol

3. Molar mass of CO2 = 12 + 16(2)

= 12 + 32

= 44g/mol

mole = 93.5/44

mole = 2.125mol

4. Molar mass of sodium nitrate (NaNO3) = 23 + 14 + 16(3)

= 23 + 14 + 48

= 85g/mol

mole = 25.6/85

mole = 0.301mol

5. Number of particles in one mole of a substance is 6.022 × 10^23 particles. Hence, in 0.75mol of calcium hydroxide (Ca(OH)2, there will be;

0.75mol × 6.02 × 10^23

= 4.515 × 10^23

= 4.52 × 10^23 particles

8 0

3 years ago

At Appliance City you sold a refrigerator to a customer for $369.00. Appliance City advertises that if a customer finds the same

Kryger [21]

Answer:

$318

Explanation:

I hope this helps!

6 0

3 years ago

Order the follow processes from (1) the least work done by the system to (5) the most work done by one mole of an ideal gas at 2

quester [9]

Answer : The order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

Explanation :

The formula used for isothermally irreversible expansion is :

$w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)$

where,

w = work done

$p_{ext}$ = external pressure

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

The expression used for work done in reversible isothermal expansion will be,

$w=-nRT\ln (\frac{V_2}{V_1})$

where,

w = work done = ?

n = number of moles of gas = 1 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = $25^oC=273+25=298K$

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

First we have to determine the work done for the following process.

(1) An isothermal expansion from 1 L to 10 L at an external pressure of 2.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(2.5atm)\times (10-1)L$

$w=-22.5L.atm=-22.5\times 101.3J=-2279.25J$

(2) A free isothermal expansion from 1 L to 100 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{100L}{1L})$

$w=-11409.6J$

(3) A reversible isothermal expansion from 0.5 L to 4 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{4L}{0.5L})$

$w=-5151.97J$

(4) A reversible isothermal expansion from 0.5 L to 40 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{40L}{0.5L})$

$w=-10856.8J$

(5) An isothermal expansion from 1 L to 100 L at an external pressure of 0.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(0.5atm)\times (100-1)L$

$w=-49.5L.atm=-49.5\times 101.3J=-5014.35J$

Thus, the order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

8 0

3 years ago