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Vilka [71]
3 years ago
12

Limestone may be made up of A) plant debris. C) fragments of granitic and andesitic rocks. 34) B) shell fragments from marine an

imals. D) animal bones.
Chemistry
1 answer:
spayn [35]3 years ago
8 0

Answer:

B

Explanation:

The debris of marine organisms such as corals, shells, and algae accumulate in the bottom of marine waters and accumulate with sediments. This debris is made up of calcium carbonate and over time the pressure and temperatures after sedimentation layers result to formation of limestone that transforms the calcium carbonate to the form of calcite.

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Which element has atoms that can bond with each other to form ring, chain, and network structures?
matrenka [14]
This is basics of organic chemistry - the element that has atoms that can bond with each other to form ring, chain, and network structures is (3) carbon.
The other elements available above cannot form such structures, whereas carbon, which is abundant in nature, can.
6 0
4 years ago
Read 2 more answers
Which bond type is found in ammonium chloride. Select one: a. metallic b. covalent c. ionic
Flura [38]

Answer:

ionic

Explanation:

In NH4Cl molecule, ionic bond is formed between NH4+ and Cl– ions, 3 covalent bonds are formed between N and three H atoms and one coordinate bond is formed between N and 1 H atom.

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7 0
3 years ago
Read 2 more answers
When 0.040 mol of propionic acid, c2h5co2h, is dissolved in 750 ml of water, the equilibrium concentration of h3o+ ions is measu
WITCHER [35]
Answer is: Ka for propanoic acid is 6,57·10⁻⁵.
Chemical reaction: C₂H₅COOH(aq) + H₂O(l) ⇄ C₂H₅COO⁻(aq) + H₃O⁺(aq).
n(C₂H₅COOH) = 0,04 mol.
V(C₂H₅COOH) = 750 mL = 0,75 L.
c(C₂H₅COOH) = 0,04 mol ÷ 0,75 L.
c(C₂H₅COOH) = 0,053 mol/L = 0,053 M.
[C₂H₅COO⁻] = [H₃O⁺] = 1,84·10⁻³ M = 0,00184 M.<span>
[HCN] = 0,053 M - 0,00184 M = 0,0515 M.
Ka = [</span>C₂H₅COO⁻] · [H₃O⁺] / [C₂H₅COOH].
Ka = (0,00184 M)² / 0,0515 M.
Ka = 6,57·10⁻⁵.
3 0
3 years ago
The energy of an electron in a multielectron atom is determined by
Wewaii [24]

Answer:

by principal quantum number (n) and azimuthal quantum number (l)

Explanation:

I used the web to answer so I'm not sure if this is right

5 0
3 years ago
The intermolecular forces present in CH 3NH 2 include which of the following? I. dipole-dipole II. ion-dipole III. dispersion IV
astra-53 [7]

Answer:

I. dipole-dipole

III. dispersion

IV. hydrogen bonding

Explanation:

Intermolecular forces are weak attraction force joining nonpolar and polar molecules together.

London Dispersion Forces are weak attraction force joining non-polar and polar molecules together. e.g O₂, H₂,N₂,Cl₂ and noble gases. The attractions here can be attributed to the fact that a non -polar molecule sometimes becomes polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant.

Dispersion forces are the weakest of all electrical forces that act between atoms and molecules. The force is responsible for liquefaction or solidification of non-polar substances such as noble gas an halogen at low temperatures.

Dipole-Dipole Attractions are forces of attraction existing between polar molecules ( unsymmetrical molecules) i.e molecules that have permanent dipoles such as HCl, CH3NH2 . Such molecules line up such that the positive pole of one molecule attracts the negative pole of another.

Dipole - Dipole attractions are more stronger than the London dispersion forces but weaker than the attraction between full charges carried by ions in ionic crystal lattice.

Hydrogen Bonding is a dipole-dipole intermolecular attraction which occurs when hydrogen is covalently bonded to highly electronegative elements such as nitrogen, oxygen or fluorine. The highly electronegative elements have very strong affinity for electrons. Hence, they attracts the shared pair of electrons in the covalent bonds towards themselves, leaving a partial positive charge on the hydrogen atom and a partial negative charge on the electronegative atom ( nitrogen in the case of CH3NH2 ) . This attractive force is know as hydrogen bonding.

7 0
4 years ago
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