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2 years ago

Name some ways of water irrigation.

Chemistry

1 answer:

Ivanshal [37]2 years ago

3 0

Answer:

Surface Irrigation. Water is scattered equally throughout the land with the help of gravitational pull and it doesn’t require a machine to take care of it.

Localized Irrigation. Water is scattered throughout the land under low pressure.

Drip Irrigation. In this process of drip irrigation, the water drops fall on the root of every plant that is around the system.

Explanation:

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tatiyna

Answer:

Experimental group

Explanation:

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4 0

3 years ago

1. How much heat (in calories) is needed to raise 20 g of H2O from 5°C to 40°C? (c = 1.0 cal/g °C)

KIM [24]

Answer:

700 calories

Explanation:

Using the formula below:

Q = m × c × ∆T

Where;

Q = amount of heat required (calories)

m = mass of substance (g)

c = specific heat of substance (cal/g°C)

∆T = change in temperature (°C)

According to this question, the following information was provided;

Q = ?

m = 20g

c = 1.0 cal/g °C

∆T = 40°C - 5°C = 35°C

Using the formula; Q = m × c × ∆T

Q = 20 × 1 × 35

Q = 700 calories

Hence, 700 cal of heat energy is needed to raise 20 g of H2O from 5°C to 40°C.

6 0

2 years ago

For an ideal gas, classify the pairs of properties as directly or inversely proportional. You are currently in a sorting module.

vredina [299]

Answer:

the result for the following are (a) P is directly proportional to n

(b) V is directly proportional to T (c) P is directly proportional to T (d) T is inversly proportional to V (e) P is inversely proportional to V

6 0

2 years ago

A closed container contains 0.40 moles of argon gas at 25 °C and a pressure of 740 torr. The container is heated to 125 °C and t

Ksivusya [100]

The number of moles of argon that must be released in order to drop.

Solution:

Initial Temperature = 25°c = 298 K

Final Temperature =125 °c = 398 K

Initial Moles (n1) = 0.40 mole

Now, Using the ideal gas law,

n1T1 = n2T2

0.400×298 = n2 × 398

n2 = 0.299 mol

Moles of Argon released

= 0.400-0.299

= 0.100 mol.

Pressure and force are related. That is using the physical equations if you know the other, you can calculate one using pressure = force/area. This pressure can be reported in pounds per square inch, psi, or Newtons per square meter N/m2. Kinetic energy causes air molecules to move faster. They hit the walls of the container more often and with greater force. The increased pressure inside the can may exceed the strength of the can and cause an explosion.

Learn more about The temperature here:-brainly.com/question/24746268

#SPJ1

7 0

1 year ago

1. A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0%

Tatiana [17]

Answer:

The mass of PbI2 will be 18.2 grams

Explanation:

Step 1: Data given

Volume solution = 99.8 mL = 0.0998 L

mass % KI = 12.0 %

Density = 1.093 g/mL

Volume of the other solution = 96.7 mL = 0.967 L

mass % of Pb(NO3)2 = 14.0 %

Density = 1.134 g/mL

Step 2: The balanced equation

Pb(NO3)2(aq) + 2 KI(aq) ⇆ PbI2(s) + 2 KNO3(aq)

Step 3: Calculate mass

Mass = density * volume

Mass KI solution = 1.093 g/mL * 99.8 mL

Mass KI solution = 109.08 grams

Mass KI solution = 109.08 grams *0.12 = 13.09 grams

Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL

Mass of Pb(NO3)2 solution = 109.66 grams

Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams

Step 4: Calculate moles

Moles = mass / molar mass

Moles KI = 13.09 grams / 166.0 g/mol

Moles KI = 0.0789 moles

Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol

Moles Pb(NO3)2 = 0.0463 moles

Step 5: Calculate the limiting reactant

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles

Step 6: Calculate moles PbI2

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2

Step 7: Calculate mass of PbI2

Mass PbI2 = moles PbI2 * molar mass PbI2

Mass PbI2 = 0.03945 moles * 461.01 g/mol

Mass PbI2 = 18.2 grams

3 0

3 years ago

Name some ways of water irrigation. ​

Name some ways of water irrigation.