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jek_recluse [69]
3 years ago
9

The freezing and thawing action of water affects a rock by?

Chemistry
2 answers:
Alex Ar [27]3 years ago
6 0
<span>The water soaks in the cracks of rocks, freezes and expands breaking the rock apart.</span>
Alexxx [7]3 years ago
4 0
The answer is letter b
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What is the instantaneous rate of formation of product C given the following information: a. Stoichiometric equation A+ B2C+ D b
quester [9]

Answer:

1.728 mol /(L*min)

Explanation:

Hello,

In the attached photo, you'll find the numerical procedure for your question.

- Take into account that the negative sign is eligible for reagents and positive for products.

Best regards!

6 0
3 years ago
How water deposits soil sediment and rock
n200080 [17]
<span>A river can only carry a load if it has adequate energy. When the energy drops below a certain level, therefore, the load is dropped. In the Thalweg (the line of fastest flow), more load is carried, and this is also where the erosion occurs, adding more load. On the inside of a meander, for example, since the Thalweg is on the outside, the velocity on the inside is very low, and so deposition occurs. On the very inside, water merely trickles past. This is incapable of transporting load, so it deposits it until it is able to carry all of it.</span>
8 0
3 years ago
NEED HELP ASAP ITS DUE TODAY THESE QUESTIONS
beks73 [17]

1.

Density can be defined as the mass of the substance in unit volume.

Density = mass / volume

Hence, g/mL and kg/L can be used as the units of density.

Those units are interchangeable because when converting one unit into other one nothing will happen to the value.

That is because when converting the g into kg, you have to multiply the value by 1 x 10⁻³. When converting mL into L, you should again multiply the volume by 1 x 10⁻³. Then those 1 x 10⁻³ will cancel off and the original value will remain as same.

2.

Answer is 2.70 g/mL.

<em>Explanation;</em>

Mass of the block = 146 g

Volume of the block =  length x width x height

                                  = 6.0 cm x 3.0 cm x 3.0 cm

                                  = 54 cm³ = 54 mL

Density = mass / volume

            = 146 g / 54 mL

            = 2.70 g/mL

3.

Answer is  3.39 g/mL.

<em>Explanation;</em>

When immersing an object in a solution, the increased volume indicates the volume of that object.

Hence,

  Volume of the object = increased volume of water level

                                      = final volume - initial volume

                                      = 27.8 mL - 21.2 mL

                                      = 6.6 mL

Mass of the object = 22.4 g

Density = Mass / Volume

            = 22.4 g / 6.6 mL

            = 3.39 g/mL

4.

Accepted value is the value that scientists and community accept as true. This is a theoretical value.

But the measured value is the value that you obtain from doing experiments. This is the actual value.

If your measured value is more close to the accepted value, then your measured value is more precise. But, if your measured value is far away from accepted value means that your value is not precise and there may have some errors.

5.

Answers : Percent error is 9.11 %

                 The element which has 7.13 g/cm³ as density is Zinc (Zn).

<em>Explanation;</em>

Percent error can be calculated by using following formula.

% error = ( (measured value - accepted value) / accepted value ) x 100%

            = ( (7.78 g/cm³ - 7.13 g/cm³) / 7.13 g/cm³ ) x 100%

            = 9.11 %

6 0
3 years ago
The decomposition of ammonia is: 2 NH3(g) ⇌ N2(g) + 3 H2(g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammoni
ValentinkaMS [17]

Answer:

"6.7\times 10^{-4} \ atm" is the right answer.

Explanation:

Given:

Partial pressure of N_2,

= 0.20 atm

Partial pressure of H_2,

= 0.15 atm

K_p = 1.5\times 10^3 at 400^{\circ} C

As we know,

⇒ K_p = \frac{pN_2\times pH_2^3}{pNH_3^2}

By putting the values, we get

    1.5\times 10^3=\frac{0.20\times (0.15)^3}{pNH_3^2}

        pNH_3^2 = \frac{0.000675}{1.5\times 10^3}

                    =6.7\times 10^{-4} \ atm

                   

3 0
2 years ago
What are the products obtained in the electrolysis of molten nai?
yanalaym [24]
Answer is: sodium (Na) and iodine (I₂).

<span> First ionic bonds in this salt are separeted because of heat: 
</span>NaI(l) → Na⁺(l) + I⁻(l).

Reaction of reduction at cathode(-): Na⁺(l) + e⁻ → Na(l) /×2.

2Na⁺(l) + 2e⁻ → 2Na(l).

Reaction of oxidation at anode(+): 2I⁻(l) → I₂(l) + 2e⁻.

The anode is positive and the cathode is negative.


4 0
3 years ago
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