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QveST [7]
3 years ago
9

Quartzite is a coarse-grained rock derived from sandstone.

Chemistry
2 answers:
TiliK225 [7]3 years ago
7 0

Answer:

granoblastic metamorphic rock

Explanation:

sashaice [31]3 years ago
6 0
Its a metamorphic rock
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If during an experiment zinc was found to be more reactive than lead or copper, zinc would be considered the strongest _____.
densk [106]

Zinc would be considered the strongest reducing agent.

<h3>Reducing agent</h3>

A reducing agent is a chemical species that "donates" one electron to another chemical species in chemistry (called the oxidizing agent, oxidant, oxidizer, or electron acceptor). Earth metals, formic acid, oxalic acid, and sulfite compounds are a few examples of common reducing agents.

Reducers have excess electrons (i.e., they are already reduced) in their pre-reaction states, whereas oxidizers do not. Usually, a reducing agent is in one of the lowest oxidation states it can be in. The oxidation state of the oxidizer drops while the oxidizer's oxidation state, which measures the amount of electron loss, increases. The agent in a redox process whose oxidation state rises, which "loses/donates electrons," which "oxidizes," and which "reduces" is known as the reducer or reducing agent.

Learn more about reducing agent here:

brainly.com/question/2890416

#SPJ4

<h3 />
5 0
2 years ago
The amount of energy that is transferred when one mole of a compound is formed from its component elements in their standard sta
amm1812
................................
5 0
3 years ago
There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In
adell [148]

Answer:

The net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide is 523.2 kJ.

Explanation:

Step 1:

CaC_2(s) + 2H_2O(g)\rightarrow C_2H_2(g) + Ca(OH)_2(s),\Delta H_1=414.0 kJ...[1]

Step 2 :

6C_2H_2(g) + 3CO_2(g) + 4H_2O(g)\rightarrow 5CH_2CHCO_2H(g) \Delta H_2=132.0kJ..[2]

Adding 6 × [1] and [2]:

6CaC_2(s) + 12H_2O(g)\rightarrow 6C_2H_2(g) + 6Ca(OH)_2(s)

6C_2H_2(g)+3CO_2(g)+16H_2O(g)\rightarrow 5CH_2CHCO_2H(g)

we get :

6CaC_2(s) + 8H_2O(g)+3CO_2(g)\rightarrow 5CH_2CHCO_2H(g)+ 6Ca(OH)_2(s),\Delta H'=?

\Delta H'=6\times \Delta H_1+\Delta H_2

\Delta H'=6\times 414.0 kJ+132.0kJ

\Delta H'=2,626 kJ

Energy released on formation of 5 moles of acrylic acid = 2,626 kJ

Energy released on formation of 1 mole of acrylic acid:

\frac{ 2,626 kJ}{5 } = 523.2 kJ

7 0
3 years ago
When methyloxirane is treated with HBr, the bromide ion attacks the less substituted position. However, when phenyloxirane is tr
konstantin123 [22]

Answer:

See explanation and picture below

Explanation:

First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.

In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.

In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.

4 0
3 years ago
On the reaction below, label the BSA, BSB, CA, and CB. CH3COOH + H2O → CH3COO– + H3O+
Ostrovityanka [42]

Answer:

Acid(BSA) = CH₃COOH

Base (BSB) = H₂O

Conjugate base (CB) = CH₃COO⁻

Conjugate acid (CA) = H₃O⁺

Explanation:

Equation of reaction;

CH₃COOH + H₂O → CH₃COO⁻ + H₃O⁺

Hello,

From my understanding of the question, we are required to identify the

1) Acid

2) Base

3) conjugate acid

4) conjugate base in the reaction

Acid (BSA) = CH₃COOH

Base (BSB) = H₂O

CA = conjugate acid = H₃O⁺

CB = conjugate base = CH₃COO⁻

7 0
3 years ago
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