What's the answers to this

When a forest burns and all that's left is some ash,

dalvyx [7]

Answer:

C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + H₂O

Explanation:

When a forest burns and all that's left is some ash, most of the mass of the trees go to the atmosphere, in the form of carbon dioxide.

In a total combustion process of organic matter, the two molecules produced are carbon dioxide and water. (CO₂ and H₂O)

The equation for the combustion of glucose is:

C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + H₂O

8 0

3 years ago

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Give an example of something that is observable but does not contain matter. Explain.

ale4655 [162]

The best example I can give that is observable but does not contain matter is:

sunlight or simply light

Light does not contain matter because it is a beam of energy or to be specific made up of packet of photons.

6 0

3 years ago

What technology can make coal a cleaner fuel? fluidized-bed combustion building smoke stacks taller implementing a prewash of co

anastassius [24]

Answer : The technology that can make coal a cleaner fuel is to use lower quality of coal with high sulfur content.

Explanation : Using lower quality of coal with high amount of sulfur in it can make coal a cleaner fuel because of high amount of sulfur it will have better combustion.

4 0

4 years ago

HELPPPPPPPPPPPPPPPPPPPP MEEEEEEEEEEEEEEEE PLZZZZZ WITHHHHH SCIENCE 13. Evaluate: What type of bonding does a material most likel

djyliett [7]

13 An Ionic Bond

these are the characteristics of an ionic bond compounds like NaCl have these characteristics

14 substances could be compounds or elements hence not all substances are compounds

15 It is not water. The chemical composition of water is H2O not H202

(H202 is hydrogen peroxide and is toxic)

6 0

3 years ago

A student mixes in a test tube 3.00mL of 0.050M CuSO4with 7.00mL of 0.20M NH3/NH41 . The solution becomes a deep blue color. Ass

valkas [14]

Answer:

$\large \boxed{\text{0.0035 mol/L}}$

Explanation:

We are given the volumes and concentrations of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble all the data in one place.

Cu²⁺ + 4NH₃ ⟶ Cu(NH₃)₄²⁺

V/mL: 3.00 7.00

c/mol·L⁻¹: 0.050 0.20

1. Identify the limiting reactant

(a) Calculate the moles of each reactant

$\text{Moles of Cu}^{2+}= \text{3.00 mL solution} \times \dfrac{\text{0.050 mmol Cu}^{2+}}{\text{1 mL solution}} = \text{0.150 mmol Cu}^{2+}\\\\\text{Moles of NH}_{3} = \text{7.00 mL solution} \times \dfrac{\text{0.20 mmol NH}_{3}}{\text{1 mL solution}} = \text{0.140 mmol NH}_{3}$

(b) Calculate the moles of Cu(NH₃)₄²⁺ that can be formed from each reactant

(i) From Cu²⁺

$\text{Moles of Cu(NH$_{3}$)$_{4}$$^{2+}$} = \text{0.150 mmol Cu}^{2+} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}}{\text{1 mmol Cu}^{2+}}\\\\= \text{0.150 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}$

(ii) From NH₃

$\text{Moles of Cu(NH$_{3}$)$_{4}$$^{2+}$} = \text{0.140 mmol NH}_{3} \times \dfrac{\text{1 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}}{\text{4 mmol NH}_{3}}\\\\= \text{0.0350 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$}$

NH₃ is the limiting reactant, because it forms fewer moles of the complex ion.

(c) Concentration of the complex ion

$\text{The reaction forms 0.0350 mmol Cu(NH$_{3}$)$_{4}$$^{2+}$ in a total volume of 10.00 mL.}\\c = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{0.0350 mmol}}{\text{10.00 mL}} = \textbf{0.0035 mol/L}\\\\\text{The concentration of the complex ion is $\large \boxed{\textbf{0.0035 mol/L}}$}$

7 0

3 years ago