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2 years ago

Compaction is most significant as a lithification process for sedimentary rocks composed of sand-sized particles. True False

Chemistry

1 answer:

user100 [1]2 years ago

8 0

Answer:

FALSE

Explanation:

Sedimentary rocks are defined as those rocks that are formed due to the combined effect of compaction and cementation process. In order to form sedimentary rocks, at first, the sediments that are transported from other places must be deposited. This deposited sediments over a due course of time, are subjected to high compaction due to the weight of the overlying weight. This results in the transformation of loose sediments into hard rocks. This how sedimentary rocks are formed which are comprised of sand-size particles. For example, Shale, Sandstone, and Mudstone.

Thus, both the compaction and cementation processes are important for the formation of sedimentary rocks.

Hence, the above-given statement is False.

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Mass in grams of 6.25 mol of copper (II) nitrate?

podryga [215]

Cu = 63.546
N= 14.001 g/mol
O= 15.999 g/mol * 3 = 47.997

Copper (II) Nitrate has a MW of 125.544 g/mol

6.25 x 125.544

= 784.65 <--- is your answer, if there were was a multiple choice or not :)

8 0

2 years ago

Read 2 more answers

Complete ionic equation K2CO3(aq)+2CuF(aq) → Cu2CO3(s)+2KF(aq) Examine each of the chemical species involved to determine the io

Fudgin [204]

Answer:

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

Explanation:

K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)

The complete ionic equation for the above equation can be written as follow:

In solution, K2CO3 and CuF will dissociate as follow:

K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)

CuF(aq) —› Ca^2+(aq) + 2F¯(aq)

Thus, we can write the complete ionic equation for the reaction as shown below:

K2CO3(aq) + 2CuF(aq) —›

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

8 0

2 years ago

A 2.0% (w/v) solution of sodium hydrogen citrate, Na2C6H6O7, which also contains 2.5% (w/v) of dextrose, C6H12O6, is used as an

tamaranim1 [39]

Answer:

0.0847M is molarity of sodium hydrogen citrate in the solution

Explanation:

The 2.0%(w/v) solution of sodium hydrogen citrate contains 2g of the solute in 100mL of solution. To find the molarity of the solution we need to convert the mass of solute to moles using molar mass and the mL of solution to Liters because molarity is the ratio between moles of sodium hydrogen citrate and liters of solution.

Moles Na2C6H6O7:

Molar Mass:

2Na: 2*22.99g/mol: 45.98g/mol

6C: 6*12.01g/mol: 72.01g/mol

6H: 6*1.008g/mol: 6.048g/mol

7O: 7*16g/mol: 112g/mol

45.98g/mol + 72.01g/mol + 6.048g/mol + 112g/mol = 236.038g/mol

Moles of 2g:

2g * (1mol / 236.038g) = 8.473x10⁻³ moles

Liters solution:

100mL * (1L / 1000mL) = 0.100L

Molarity:

8.473x10⁻³ moles / 0.100L =

<h3>0.0847M is molarity of sodium hydrogen citrate in the solution</h3>

3 0

1 year ago

Read 2 more answers

The reaction of 0.779 g K with O2 forms 1.417 g potassium superoxide, a substance used in self-contained breathing devices. Dete

zhannawk [14.2K]

First we need to find the number of moles of both K and O reacted
K - 0.779 g / 39 g/mol
= 0.02 mol
the mass of O₂ reacted = 1.417 g - 0.779 g = 0.638 g
O₂ moles = 0.638 g / 32 g/mol
= 0.02 mol
the number of both K and O₂ moles reacted are equal
therefore stoichiometry of K to O₂ reacted are 1:1
then the formula of potassium superoxide is KO₂

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2 years ago

list three sources of error that could account for the differences between your values for the enthalpy of fusion of water and t

NARA [144]

Three sources of error that might account for the differences in the enthalpy of fusion include the room temperature how much’ long you stirred and another thing that might make it have different results is how long the ice was out for

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2 years ago