Arrange the following elements in order of decreasing ionization energy: Cl, Si, P, Ar *

What are the processes for the 3 types of weathering

sweet [91]

The processes for the 3 types of weathering are,
biological: rocks worn down by plants
physical: water freezing and thawing in cracks of rocks
chemical: Acid rain
Hope this helped!!

7 0

3 years ago

PLEASE SOMEONE HELP ME WITH THIIS QUESTION

sdas [7]

Answer:

The correct answer is - 5 carbon compounds due to low to high intermolecular forces between their molecules.

Explanation:

Bottle C has gas in it and we know that alkane has carbon and hydrogen only which means they have a single sigma bond between them and very low intermolecular forces in between molecules and are present mostly at gaseous state. Thus, bottle C has alkane.

Alcohols have -OH group that can form rarely two pi bonds which means they have intermediate intermolecular force whereas acids have -cooH group with a high molecular force so bottle B with liquid is alcohol and A has acid.

3 0

3 years ago

Name: _________________ Temperature o fwater_25_degreecent.YOU MUST SHOW ALL CALCULATIONS TO RECEIVE CREDIT FOR THEM! DATA ANALY

Anton [14]

Answer:

1. 0.02 M

2. 0.01 M

3. 4×10⁻⁶

Explanation:

We know that V₁S₁ = V₂S₂

1.

Concentration of HCl = 0.05 M

end point comes at = 10 ml

So, concentration of OH⁻(aq) = [OH⁻(aq)] ⇒ (0.05 × 10) ÷ 25 ⇒ 0.02 M

2.

2mol of OH⁻(aq) ≡ 1 mole of Ca²⁺(aq)

[Ca²⁺] = 0.02 ÷ 2 = 0.01 M

3.

$K_{sp}$ = [Ca²⁺(aq)] [OH⁻(aq)]²

Ca(OH)₂ (aq) ⇄ Ca²⁺ (aq) + 2OH⁻ (aq)

$K_{sp}$ = [0.01 × (0.02)²] = 4×10⁻⁶

4.

If reaction is exothermic which means heat energy will get evolved as a result temperature of the reaction media will get increased during the course of the reaction. If temperature is externally increased, the reaction will go backward to accumulate extra heat energy.

5.

$K_{sp}$ value describes the solubility of a particular ionic compound. The higher the $K_{sp}$ value, the higher the Solubility will be.

6.

This may be due to uncommon ion effect. The process of other ions (K⁺ or Na⁺) may increase the solubility

4 0

3 years ago

Calculate the mass (in grams) of 250mL of ether at 25 oC. The density of

leonid [27]

Volume=250mL
Density=0.71g/ml

$\boxed{\sf Density=\dfrac{Mass}{Volume}}$

$\\ \sf{:}\implies Mass=Density(Volume)$

$\\ \sf{:}\implies Mass=0.71(250)$

$\\ \sf{:}\implies Mass=177.5g$

6 0

2 years ago

A 0.964 gram sample of a mixture of sodium formate and sodium chloride is analyzed by adding sulfuric acid. The equation for the

frosja888 [35]

Answer: 67.8 %.

Explanation:

Okay, let us delve right into the solution to the question;

The balanced chemical reaction is given by the equation (1) below;

2 HCOONa + H2SO4 ---------> 2 CO + 2 H2O + Na2SO4. ----------------------------------------------------------------------------(1).

From the balanced chemical reaction in equation (1) above we can see that; 2 moles of HCOONa reacts with one moles of tetraoxosulphate acid, H2SO4 to produce 2 moles of carbonmonoxide,CO; 2 moles of water, H2O and 1 mole of sodium tetraoxosulphate, Na2SO4.

The parameters given from the question are; total atmospheric pressure, P(t) = 752 torr, volume of CO= 242 mL = 0.242 Litres.

STEP ONE : find the carbon monoxide,CO pressure; P(CO).

Using the formula below;

P(t) = P(CO) + P(H2O). Hence;

P(CO) = P(t) - P(H2O). Note that P(H2O)= 19.8 torr.

==>P(CO)= 752 torr - 19.8 torr = 732.2 torr.

STEP TWO: calculate the number of moles of Carbonmonoxide,CO.

Using the formula below;

Number of moles= pressure(P) × volume(v) / gas constant(R) × temperature (T).

That is, n= PV/RT.

n= 732 torr × 0.242 Litres/ 62.4 × 295.15.

= 9.62 × 10^-3 mol of CO.

STEP THREE:

2 moles of HCOONa = 2 moles of CO.

=> 2 moles of HCOONa = 2 moles of CO/ 2 moles of CO = 1 mol( HCOONa/ CO).

Then, 9.62 × 10^-3 mol of CO × 1 mol( HCOONa/ CO).

==> 9.62 × 10^-3 mol HCOONa × molar masss of HCOONa(68 grams/mol)

= 0.654 grams.

Therefore, the percentage of sodium formate in the original mixture = 0.654 grams/ 0.964 gram × 100 = 67.8 %.

5 0

3 years ago