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djyliett [7]
3 years ago
10

Some soils, like clay, hold water tightly. Other soils, like sand, let the water run through. What property of soils is MOST IMP

ORTANT in determining how water moves through the soil?
Chemistry
1 answer:
RoseWind [281]3 years ago
8 0
Most likely the second one (sand letting water run through it) because if you are trying to determine how what moves through soil You would want the one the one that can do that:)
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5 0
3 years ago
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What is the final temperature of a 34.2 g of water initially at 282 K that has been heated with 2.71 kJ of energy?
lana66690 [7]

Answer: The final temperature of copper is  

Further explanation:

The property is a unique feature of the substance that differentiates it from the other substances. It is classified into two types:

1. Intensive properties:

These are the properties that depend on the nature of the substance. These don't depend on the size of the system. Their values remain unaltered even if the system is further divided into a number of subsystems. Temperature, refractive index, concentration, pressure, and density are some of the examples of intensive properties.

2. Extensive properties:

These are the properties that depend on the amount of the substance. These are additive in nature when a single system is divided into many subsystems. Mass, enthalpy, volume, energy, size, weight, and length are some of the examples of extensive properties.

Specific heat is the amount of heat required to increase the temperature of any substance per unit mass. Specific heat capacity is also known as mass specific heat. Its SI unit is Joule (J).

The formula to calculate the heat energy of copper is as follows:

                                       …… (1)

Here,

Q is the amount of heat transferred.

m is the mass of copper.

c is the specific heat of copper.

is the change in temperature of copper.

Rearrange equation (1) to calculate the temperature change.

                                   …… (2)

The value of Q needs to be converted into J. The conversion factor for this is,

So the value of Q can b calculated as follows:

The value of Q is 4689 J.

The value of m is 34.2 g.

The value of c is .

Substitute these values in equation (2).

The temperature change  can be calculated as follows:

                         …… (3)

Here,

is the change in temperature.

is the final temperature.

is the initial temperature.

Rearrange equation (3) to calculate the final temperature.

                      …… (4)

The value of  is .

The value of  is  

Substitute these values in equation (4).

So the final temperature of copper is .

7 0
2 years ago
What is neutron number for calcium ​
notka56 [123]

Answer:

20 neutrons

Explanation:

(not really any just look at a periodic table)

5 0
2 years ago
How can an invasive species like weeds affect crops on a farm?
Hoochie [10]

Answer:

E

Explanation:

All of the above

3 0
3 years ago
A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this mat
Lady_Fox [76]

Explanation:

(a)  The given data is as follows.  

         Load applied (P) = 1000 kg

         Indentation produced (d) = 2.50 mm

         BHI diameter (D) = 10 mm

Expression for Brinell Hardness is as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}    

Now, putting the given values into the above formula as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}                            = \frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}  

                       = \frac{2000}{9.98}                          

                       = 200

Therefore, the Brinell HArdness is 200.

(b)     The given data is as follows.

               Brinell Hardness = 300

                Load (P) = 500 kg

               BHI diameter (D) = 10 mm

             Indentation produced (d) = ?

                      d = \sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}

                         = \sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}

                          = 4.46 mm

Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.

5 0
2 years ago
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