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3 years ago

Soil horizons develop as a result of

Chemistry

2 answers:

iris [78.8K]3 years ago

7 0

Answer:

WVCA - Soil Formation and Soil Horizons. Soils develop as a result of the interactions of climate, living organisms, and landscape position as they influence parent material decomposition over time.

Explanation:

harkovskaia [24]3 years ago

4 0

The interactions of climate, living organisms, and landscape .

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The bond enthalpy of the Brâ€“Cl bond is equal to Î”HÂ°rxn for the following reaction. BrCl(g) â†’ Br(g) + Cl(g) Using the follo

ddd [48]

Answer:

a. Br2 (l) â†’ Br2(g) Î”HÂ°rxn = 30.91 kJ/mol

Explanation:

Bond enthalpy is the energy which is required to break 1 mol of of bonds in gaseous covalent molecules. Bond breaking process can be endothermic or exothermic depending on the molecules association. The process needs to be performed in chemical laboratory under standard conditions.

3 0

3 years ago

What is the density (in g/L) of a gas with a molar mass of 16.01 g/mol at 1.75 ATM and 337 K?

IRINA_888 [86]

You can use this formula to solve for density--> Density= PM/ RT, where P is pressure, M is molar mass, R is the gas constant and T is temperature.

P= 1.75 atm
M= 16.01 g/ mol
R= 0.0821 atm·L/ mol·K
T=337 k

density= (1.75 x 16.01)/ (0.0821 x 337)= 1.01 g/L

8 0

4 years ago

Read 2 more answers

A chemist needs to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solutio

Zigmanuir [339]

Answer: The concentration of original sulfuric acid solution is 1.62 M

Explanation:

Let the original concentration of sulfuric acid be 'x' M

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated sulfuric acid solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted sulfuric acid solution

We are given:

$M_1=xM\\V_1=25.00mL\\M_2=?M\\V_2=250.0mL$

Putting values in above equation, we get:

$x\times 25.00=M_2\times 250.0\\\\M_2=\frac{x\times 25.0}{250}=\frac{x}{10}$

Now, to calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=\frac{x}{10}M\\V_1=10.00mL\\n_2=1\\M_2=0.1790M\\V_2=18.07mL$

Putting values in above equation, we get:

$2\times \frac{x}{10}\times 10.00=1\times 0.1790\times 18.07\\\\x=\frac{1\times 0.1790\times 18.07\times 10}{2\times 10.00}=1.62M$

Hence, the concentration of original sulfuric acid solution is 1.62 M

5 0

4 years ago

0005x0.002 into scientific notation

geniusboy [140]

0005 * 0.002 = 5 * 0.002 = 0.010 = 10^-2

7 0

3 years ago

The two common chlorides of phosphorus, PCl3, and PCl5, both important for the production of the other phosphorus compounds, coe

Irina18 [472]

Answer:

For a: The value of $K_c$ for the given reaction is 271.6

For b: The value of $K_p$ for the reaction is 6.32

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $PCl_5$ :

Given mass of $PCl_5$ = 0.105 g

Molar mass of $PCl_5$ = 208.24 g/mol

Putting values in equation 1, we get:

$\text{Moles of }PCl_5=\frac{0.105g}{208.24g/mol}=5.04\times 10^{-4}mol$

For $PCl_3$ :

Given mass of $PCl_3$ = 0.220 g

Molar mass of $PCl_5$ = 137.33 g/mol

Putting values in equation 1, we get:

$\text{Moles of }PCl_3=\frac{0.220g}{137.33g/mol}=1.60\times 10^{-3}mol$

For $Cl_2$ :

Given mass of $Cl_2$ = 2.12 g

Molar mass of $Cl_2$ = 71.0 g/mol

Putting values in equation 1, we get:

$\text{Moles of }Cl_2=\frac{2.12g}{71.0g/mol}=0.029mol$

Volume of the flask = 25.0 L

For the given chemical equation:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

For a:

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

The expression of $K_c$ for above reaction follows:

$K_c=\frac{PCl_5}{PCl_3\times Cl_2}$

We are given:

$[PCl_5]=\frac{5.04\times 10^{-4}mol}{25L}$

$[PCl_3]=\frac{1.60\times 10^{-3}mol}{25L}$

$[Cl_2]=\frac{0.029mol}{25L}$

Putting values in above equation, we get:

$K_c=\frac{(\frac{5.04\times 10^{-4}}{25})}{(\frac{1.60\times 10^{-3}}{25})\times (\frac{0.029}{25})}\\\\K_c=271.6$

Hence, the value of $K_c$ for the given reaction is 271.6

For b:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = 271.6

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $250^oC=250+273=523K$

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=1-2=-1$

Putting values in above equation, we get:

$K_p=271.6\times (0.0821\times 523)^{-1}\\\\K_p=6.32$

Hence, the value of $K_p$ for the reaction is 6.32

7 0

4 years ago