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Natali [406]
3 years ago
6

What type of organic compound has the empirical formula CH2O and can serve as energy storage or the starting material for fats a

nd amino acids?
Amine
Carbohydrate
Carbonyl
Ketone
*please explain your answer for brainliest
Chemistry
1 answer:
Luda [366]3 years ago
4 0
Carbohydrates. CH2O is the most basic formula of simple sugars.Empirical formula is the lowest whole number ratio of atoms in a formula.For example glucose is C6H12O6 can be reduced by 6 to get CH2O. Simples sugars include glucose and fructose which can serve as sources of energy for fats and amino acids.
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If a farmer wants to increase the health of his crops, he could apply a nitrogen-based fertilizer. why is nitrogen important to
Finger [1]
Not only for plants, but also for every organism and every life form nitrogen is important because  it poses as catalyst that supports different is chemical reactions of growth. N is filled with chlorophyll that actually gives you the answer on the question as chlorophyll allow plant to perform photosynthesis by absorbing energy from light that leads to the growth. So, it is important because it provides <span>photosynthesis.</span>
8 0
3 years ago
What is a possible cause of a large percentage of error in an experiment where mgo is produced from the combustion of magnesium?
PilotLPTM [1.2K]

A possible cause of a large percentage of error in an experiment where MgO is produced from the combustion of magnesium would be not all of the Mg has completely reacted. <span>

I hope this helps and if you have any further questions, please don’t hesitate to ask again. </span>

4 0
3 years ago
Read 2 more answers
A gas system has an initial number of moles of 0.693 moles with the volume unknown. When the number of moles changes to 0.928 mo
lara [203]

Answer:

The initial volume in mL is 5959.2 mL

Explanation:

As the number of moles of a gas increases, the volume also increases.  Hence, number of moles and volumes are directly proportional i.e

n ∝ V

Where n is the number of moles and V is the volume

Then, n = cV

c is the proportionality constant

∴n/V = c

Hence n₁/V₁ = n₂/V₂

Where n₁ is the initial number of moles

V₁ is the initial volume

n₂ is the final number of moles

and V₂ is the final volume.

From the question,

n₁ = 0.693 moles

V₁ = ?

n₂ = 0.928 moles

V₂ = 7.98 L

Putting the values into the equation

n₁/V₁ = n₂/V₂

0.693 / V₁ = 0.928 / 7.98

Cross multiply

∴ 0.928V₁ = 0.693 × 7.98  

0.928V₁ = 5.53014

V₁ = 5.53014/0.928

V₁ = 5.9592 L

To convert to mL, multiply by 1000

∴ V₁ = 5.9592 × 1000 mL

V₁ = 5959.2 mL

Hence, the initial volume in mL is 5959.2 mL

5 0
3 years ago
How many moles of ethanol are produced starting with 500.g glucose?
Monica [59]
<h3>Answer:</h3>

5.55 mol C₂H₅OH

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Tables
  • Moles

<u>Stoichiometry</u>

  • Using Dimensional Analysis
  • Analyzing Reactions RxN
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

[Given] 500. g C₆H₁₂O₆ (Glucose)

[Solve] moles C₂H₅OH (Ethanol)

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH

[PT] Molar mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

<u>Step 3: Stoichiometry</u>

  1. [DA] Set up conversion:                                                                                 \displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})
  2. [DA} Multiply/Divide [Cancel out units]:                                                         \displaystyle 5.55001 \ mol \ C_2H_5OH

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH

8 0
3 years ago
An acid with the general formula RCOOH is used to make a 0.10 M solution in water. The pH of this acid solution is measured as 3
Sati [7]

Answer:

0.159 \%

Explanation:

The acid will dissociate according to the reaction shown below:-

RCOOH + H_2O\rightleftharpoons RCOO^- + H_3O^+

Given that, pH=3.8

The concentration of can be determined from the expression fo pH as:-

pH = - log [H^+]

3.8  = - log [H^+]

[H^+] = 1.59\times 10^{-4}\ M

The initial concentration of RCOOH was 0.10 M, then the percent dissociation was- calculated as shown below:-

\% \ dissociation=\frac{1.59\times 10^{-4}}{0.10}\times 100=0.159\ \%

3 0
3 years ago
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