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zysi [14]
2 years ago
13

Samples of this drawing medium have been dated back to 30,000 BCE. Group of answer choices silverpoint charcoal pencil color pen

cil pastel
Chemistry
1 answer:
GuDViN [60]2 years ago
7 0

Charcoal is the sample of this drawing medium have been dated back to 30,000 BCE and is denoted as option B.

<h3>What is Charcoal?</h3>

This is referred to a black carbon residue which is heated in insufficient oxygen to give it a light weight.

It is used for drawing and has been in existence for as long as 30,000 BCE hence the reason why it is the most appropriate choice.

Read more about Charcoal here brainly.com/question/22220547

#SPJ1

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PLEASE HELP! 25 POINTS!!!! I got the #1, just not #2 and #3. An industrial chemical company has opened a new plant that will pro
podryga [215]

Answer :

Part 1 : Balanced reaction, 3H_2(g)+N_2(g)\rightarrow 2NH_3(g)

Part 2 : The theoretical yield of NH_3 gas = 440.96 g

Part 3 : The % yield of ammonia is 90.03 %

Solution : Given,

Mass of N_2 = 475 g

Molar mass of N_2 = 28 g/mole

Molar mass of NH_3 = 17 g/mole

Experimental yield of NH_3 = 397 g

<u>Answer for Part (1) :</u>

The balanced chemical reaction is,

3H_2(g)+N_2(g)\rightarrow 2NH_3(g)

<u>Answer for Part (2) :</u>

First we have to calculate the moles of N_2.

\text{Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molecular mass of }N_2}=\frac{475g}{28g/mole}=16.96moles

From the given reaction, we conclude that

1 moles of N_2 gas react to give 2 moles of NH_3 gas

16.96 moles of N_2 gas react to give \frac{2}{1}\times 16.96=33.92 moles of NH_3 gas

Now we have to calculate the mass of NH_3 gas.

\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3

\text{ Mass of }NH_3=(33.92moles)\times (17g/mole)=440.96g

Therefore, the theoretical yield of NH_3 gas = 440.96 g

<u>Answer for Part (3) :</u>

Formula used for percent yield :

\% \text{ yield of }NH_3=\frac{\text{ Experimental yield of }NH_3}{\text{ Theoretical yield of }NH_3}\times 100

\% \text{ yield of }NH_3=\frac{397g}{440.96g}\times 100=90.03\%

Therefore, the % yield of ammonia is 90.03 %

3 0
2 years ago
PLEASE AWNSER THIS ASAP!!! 100points! i will give brainliest please awnser right!!!
GaryK [48]

Answer: How does cellar respiration the law of converter of mass and the law of conversion of energy?

Explanation:  Conservation of mass is the first law of thermodynamics. This states that energy is conserved in all processes, since it cannot be created nor destroyed. This applies to photosynthesis and cellular respiration.

8 0
2 years ago
Read 2 more answers
On the ph scale, indicate which direction is increasingly acidic and which is increasingly basic.
alexdok [17]
1 2 3 4 5 6 7 8 9 10 11 12 13 14
From 1-6.5 ph acidic
From 7-14 ph alkaline
8 0
2 years ago
Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,
WITCHER [35]

Answer: Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]

We are given:

\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ

Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

6 0
2 years ago
When the following equation is balanced using the smallest possible integers, what is the number in front of the substance in bo
weeeeeb [17]

The balanced equation :

8Al + 3Fe₃O₄→ 4Al₂O₃ + 9Fe

You just have to look for which element is in bold because the question is not clear

<h3>Further explanation</h3>

Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:

1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.

2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product

3. Select the coefficient of the substance with the most complex chemical formula equal to 1

Reaction

Al + Fe₃O₄→ Al₂O₃ + Fe

give a coefficient

aAl + Fe₃O₄→ bAl₂O₃ + cFe

O, left=4, right=3b⇒3b=4⇒b=4/3

Al, left=a, right=2b⇒a=2b⇒a=2.4/3⇒a=8/3

Fe, left=3, right=c⇒c=3

The equation becomes :

8/3Al + Fe₃O₄→ 4/3Al₂O₃ + 3Fe x3

8Al + 3Fe₃O₄→ 4Al₂O₃ + 9Fe

7 0
3 years ago
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