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nikitadnepr [17]
3 years ago
9

Can you please help me with the green question

Chemistry
1 answer:
solmaris [256]3 years ago
8 0
I was flung towards the atmosphere. Slowly but surely, I made my way inside. On my way there, I began to heat up. I began to vibrate rapidly, and I noted it happened as I entered the atmosphere. I think it was because we sped up so much, but I can't say for sure because I'm an air particle... However, I do know that I am now a heated air particle due to friction and movement from entering the atmosphere. I think I'll cool down on my descent, I don't know that I can get much hotter than this.
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How do living and nonliving things differ?
Nataly [62]

Answer:

Hey mate.....

Explanation:

This is your answer,

Living things differ from non living things through motion, action, expression, way of living and etc......

Livings things:-

1. Livings things means something that keeps on moving or doing any particular work.

2. Livings things mostly consume, remove, and reproduce through different ways.

3. Examples of livings things are humans, animals, birds, reptiles, and even the germs like bacteria, fungi, ameoba and etc....

Non Livings things:-

1. Non Living things means something that just stays like a statue usually and never move or do a particular work.

2. Non living things do not consume, remove or reproduce anything.

3. Examples of non living things are table, chair, fan, house, book etc....

hope it helps you mate....

MARK ME AS THE BRAINLIEST.....

Follow me.....!!!!

5 0
3 years ago
The combining of the nuclei of atoms is known as nuclear _____.
tankabanditka [31]
The answer is B.) fusion
7 0
3 years ago
Read 2 more answers
Suppose a student started with 142.0 mg of trans-cinnamic acid, 412 mg of pyridinium tribromide, and 2.30 mL of glacial acetic a
nirvana33 [79]

Answer: Theoretical Yield = 0.2952 g

               Percentage Yield = 75.3%

Explanation:

Calculation of limiting reactant:

n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol

pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

  • n-trans-cinnamic acid is the limiting reactant

The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

                             =  (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g

Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

                                                                                           = 75.3%

4 0
4 years ago
12. How many molecules of glucose, C6H12O6, are present in a 152 g sample
ELEN [110]

Q.No. 12:

Answer:

                 5.08 × 10²³ Glucose Molecules

Solution:

Data Given:

                Mass of Glucose  =  152 g

                M.Mass of Glucose  =  180.156 g.mol⁻¹

Step 1: Calculate Moles of Glucose as,

                Moles  =  Mass ÷ M.Mass

Putting values,

                Moles  =  152 g ÷ 180.156 g.mol⁻¹

                Moles  =  0.8437 mol

Step 2: Calculate number of Glucose Molecules,

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of glucose molecules can be written as,

 Moles  =  Number of Glucose Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of Glucose Molecules,

 Number of Glucose Molecules  =  Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting value of moles,

 Number of Glucose Molecules  =  0.8437 mol × 6.022 × 10²³ Atoms.mol⁻¹

 Number of Glucose Molecules =  5.08 × 10²³ Glucose Molecules

______________________________________________

Q.No. 12: (A)

Answer:

                3.04 × 10²⁴ Carbon Atoms

Solution:

              The molecular formula of Glucose is C₆H₁₂O₆. This specifies that there are six carbon atoms in one molecule of Glucose.

Hence,  when,

               1 molecule of Glucose contain  =  6 atoms of Carbon

Then,

     5.08 × 10²³ Glucose Molecules will contain  =  X atoms of Carbon

Solving for X,

                     X =  5.08 × 10²³ molecules  × 6 atoms / 1 molecule

                     X  =  3.04 × 10²⁴ Carbon Atoms

______________________________________________

Q.No. 12: (B)

Answer:

                1.22 × 10²⁵ Atoms in total

Solution:

              The molecular formula of Glucose is C₆H₁₂O₆. This specifies that there are 6 carbon atoms, 12 hydrogen atoms and 6 oxygen atoms in one molecule of Glucose. So, there are 24 atoms in one molecule of glucose

Hence,  when,

                    1 molecule of Glucose contain  =  24 atoms

Then,

          5.08 × 10²³ Glucose Molecules will contain  =  X atoms

Solving for X,

                     X =  5.08 × 10²³ molecules  × 24 atoms / 1 molecule

                     X  =  1.22 × 10²⁵ Atoms in total

______________________________________________

Q. No. 13

Answer:

                   1061.81 g of Aluminium

Solution:

Data Given:

                Number of F.Units  =  2.37 × 10²⁵

                A.Mass of Aluminium  =  26.98 g.mol⁻¹

                Mass of Aluminium  =  ?

Step 1: Calculate Moles of Aluminium,

                  Moles  =  Number of F.Units ÷ 6.022 × 10²³ F.Units.mol⁻¹

Putting value,

                  Moles  =   2.37 × 10²⁵ F.Units ÷ 6.022 × 10²³ F.Units.mol⁻¹

                  Moles  =  39.35 mol

Step 2: Calculate Mass of Aluminium as:

                  Moles  =  Mass ÷ A.Mass

Solving for Mass,

                  Mass  =  Moles × A.Mass

Putting values,

                  Mass  =  39.35 mol × 26.98 g.mol⁻¹

                  Mass  =  1061.81 g of Aluminium

8 0
3 years ago
water is considered as a compound of hydrogen and oxygen and not a mixture of hydrogen and oxygen and comment on it
HACTEHA [7]
Wat is considered a compound of hydrogen and oxygen and not a mixture because hydrogen and oxygen have gone through a chemical process to form the compound h2O. A mixture can generally be easily separated whereas a compound must undergo another chemical process to separate its elements. Therefore to separate oxygen and hydrogen from water a chemical process is necessary. 
4 0
4 years ago
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