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Genrish500 [490]
3 years ago
7

A___________ consists of a mixture of graphite powder (a form of carbon) and clay that is baked and hardened and encased in wood

or paper. a. silverpoint b. pencil c. charcoal d. chalk e. paste
Chemistry
1 answer:
Harlamova29_29 [7]3 years ago
5 0

Answer:

A_____PENCIL______ consists of a mixture of graphite powder (a form of carbon) and clay that is baked and hardened and encased in wood or paper. a. silverpoint b. pencil c. charcoal d. chalk e. paste

Explanation:

A pencil is a writing material, where the graphite is what gives the black color to the writing stroke.

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5. Name five plants and animals that lived in ancient Antarctica
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Answer:

Two hundred million years ago, Antarctica was a lush, temperate rainforest home to crocodile-sized amphibians and rhinoceros-sized dinosaurs.

Explanation:

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1 ankylosaurs (the armored dinosaurs), mosasaurs and plesiosaurs (both marine reptilian groups).

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8 0
2 years ago
The following reaction has been reported in the chemical literature and gives a single organic product in high yield. Write the
Alla [95]

Answer:

Explanation:

The principle applied is the Markovnikoff's rule which states that when hydrogen chloride adds to a double bond, the hydrogen atoms join to the carbon that already has the most hydrogen atoms bonded to it. The rule wa postulated by a russian chemist known as Vladimir Markovnikoff.

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8 0
3 years ago
Problem Page Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.2 g o
stich3 [128]

Answer: 0.0 grams

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of butane

\text{Number of moles}=\frac{5.2g}{58.12g/mol}=0.09moles

b) moles of oxygen

\text{Number of moles}=\frac{32.6g}{32g/mol}=1.02moles

2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O

According to stoichiometry :

2 moles of butane require 13 moles of O_2

Thus 0.09 moles of butane will require =\frac{13}{2}\times 0.09=0.585moles  of O_2

Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.

Thus all the butane will be consumed and 0.0 grams of butane will be left.

7 0
3 years ago
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