Which material typically make up the A horizon in soil

1 answer:

3 0

<span> A horizon is a surface horizon that largely consists of minerals which holds sand, silt, and clay)and with appreciable amounts of organic matter. This horizon ison top layer of many soils in grasslands and agricultural lands.</span>

You might be interested in

Please help asap

Helen [10]

C. carbon
sodium is an
oxygen is o
cl is chlorine

4 0

3 years ago

The word weather is most commonly used to refer to the state of the atmosphere at a certain time. Why do you think the same word

s344n2d4d5 [400]

<span> weather is the condition of the climate in a particular place at a particular
time.</span>

3 0

3 years ago

True or false. adding more neutrons to a neutrally charged atom will not change the charge

san4es73 [151]

The statement is true.

6 0

3 years ago

(d) carbon monoxide

Irina18 [472]

Answer:

D. carbon monoxide

Hope this helps :)

6 0

2 years ago

The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh

Wewaii [24]

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

$3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ$

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

$Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ$

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

$150.0 mL \times \frac{1.02g}{mL} = 153 g$

Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

$Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J }{\frac{4.184J}{g.\° C } \times 153g} = 5.87 \° C$

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

Learn more: brainly.com/question/4400908

7 0

2 years ago