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meriva
4 years ago
11

Which factors affect a river's load?

Chemistry
2 answers:
Tresset [83]4 years ago
7 0

Answer:

Option (D)

Explanation:

The river load is defined as the materials such as rock fragments, stones, sediments and other types of particles, which are carried by the river during its flowing stage. These particles are generally considered as the load.

A load of a river is often affected by the slope of the river, the shape of the river bed and the volume of water flow.

The slope of the river plays an essential role as the higher the slope, the higher will be the flow rate and more amount of sediments and rock fragments will be transported by the river.

The narrow type of rivers are generally faster and carries a larger amount of sediments in comparison to a broad river channel. Again, the more the volume of water flow in any river, the more it has the capacity to pick sediments and transports them from one place to another.

The lighter particles generally flow in suspension or solution, and the coarser particles are carried either by rolling, sliding and saltation process.

Hence, the correct answer is option (D).

Marizza181 [45]4 years ago
4 0
To answer the question that is: "which factors affect a river's load", we have to understand that all the things mentioned are important. All these options (river's slope, streambed shape and volume of flow) affect the amount of energy that the river has to spend and the way the river spend that energy, so, it is right to mark the alternative <span>d) all of the above.</span>
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(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct
Zarrin [17]

Explanation:

The given data is as follows.

   \Lambda^{o}_{m}(NaCl) = 1.264 \times 10^{-2}

   \Lambda^{o}_{m}(H-O=C-ONO) = 1.046 \times 10^{-2}

   \Lambda^{o}_{m}(HCl) = 4.261 \times 10^{-2}

Conductivity of monobasic acid is 5.07 \times 10^{-2} S m^{-1}

     Concentration = 0.01 mol/dm^{3}

Therefore, molar conductivity (\Lambda_{m}) of monobasic acid is calculated as follows.

                 \Lambda_{m} = \frac{conductivity}{concentration}

                                  = \frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}

                                 = \frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}

                                 = 5.07 \times 10^{-3} S m^{2} mol^{-1}

Also, \Lambda^{o}_{m} = \Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}

                            = 4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}

                            = 4.043 \times 10^{-2} S m^{2} mol^{-1}

Relation between degree of dissociation and molar conductivity is as follows.

               \alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}

                             = \frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}

                             = 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

Putting the values into the above formula we get the following.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

                        = \frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}

                        = 0.017973 \times 10^{-2}

                       = 1.7973 \times 10^{-4}

Hence, the acid dissociation constant is 1.7973 \times 10^{-4}.

Also, relation between pK_{a} and K_{a} is as follows.

                 pK_{a} = -log K_{a}

                              = -log (1.7973 \times 10^{-4})

                              = 3.7454

Therefore, value of pK_{a} is 3.7454.

                             

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lozanna [386]

Answer: yes even temperature too in rarely case

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Answer:

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cestrela7 [59]
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