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sergejj [24]
2 years ago
8

Maple syrup, which comes from the sap of maple trees, contains water and natural sugars. It's a clear, brown liquid and the suga

rs can’t be separated by filtration. Under which category will you classify it?
Chemistry
1 answer:
il63 [147K]2 years ago
6 0
Maple syrup would be categorized under solutions. As it contains sugar which is very soluble in water, it dissolves completely in water depending on its solubility at that temperature making it impossible for the components to be filtered out. A solution is a homogeneous mixture.
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What is the maximum amount in moles of P2O5P2O5 that can theoretically be made from 112 gg of O2O2 and excess phosphorus
Nastasia [14]

Answer:

n_{P_2O_5}^{max}=1.4molP_2O_5

Explanation:

Hello!

In this case, since the reaction between phosphorous and oxygen to form diphosphorous pentoxide is:

2P+\frac{5}{2}O_2\rightarrow P_2O_5

Thus, since phosphorous is in excess and oxygen and diphosphorous pentoxide are in a 5/2:1 mole ratio, we can compute the maximum moles of product as shown below:

n_{P_2O_5}^{max}=112 gO_2*\frac{1molO_2}{32.00gO_2}*\frac{1molP_2O_5}{5/2molO_2}\\\\  n_{P_2O_5}^{max}=1.4molP_2O_5

Best regards!

5 0
2 years ago
The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make
natta225 [31]

Answer:

the energy of the third excited rotational state \mathbf{E_3 = 16.041 \ meV}

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = \dfrac{m_1 \times m_2}{m_1 + m_2}

μ = \dfrac{1 \times 35}{1 + 35}

μ = \dfrac{35}{36}

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ  = \\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \  kg

μ  = 1.6139 × 10⁻²⁷ kg

r_o = 127 \ pm = 127*10^{-12} \ m

The rotational level Energy can be expressed by the equation:

E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)

where ;

J = 3 ( i.e third excited state)  &

I = \mu r^2_o

E_J= \dfrac{h^2}{8  \pi  \mu r^ 2 \mur_o } \times J ( J +1)

E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8  \times  \pi ^2  \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2  } \times 3 ( 3 +1)

E_3= 2.5665 \times 10^{-21} \ J

We know that :

1 J = \dfrac{1}{1.6 \times 10^{-19}}eV

E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV

E_3 = 16.041  \times 10 ^{-3} \ eV

\mathbf{E_3 = 16.041 \ meV}

8 0
2 years ago
4 Weathering Assessment
Leni [432]

Answer:

This is an example of a physical change because the ice cubes began to melt.

This is an example of a physical change because the ice cubes began to melt.

Explanation:

The above is the right answer to the question about the dissolution of the whole mixture mentioned in the excerpts above.

6 0
3 years ago
For the following reaction, identify the element that was oxidized, the element that was reduced, and the reducing agent. Give a
Volgvan

<span>For this reaction, oxidation number of Carbon in CO would be +2 while oxidation number of carbon in CO2 would be +4 and so this means that carbon has oxidized. Oxidation number of nitrogen in NO is +2. While oxidation number of nitrogen in N2 is 0 so this means that nitrogen had reduced. The reducing agent is the one which provides electrons by oxidizing itself so in this case; CO is the reducing agent while the C in CO oxidized to produce electrons. </span><span>I am hoping that this answer has satisfied your query about and it will be able to help you, and if you’d like, feel free to ask another question.</span>

6 0
3 years ago
A sample of CO2 weighing 86.34g contains how many molecules?
irakobra [83]

Answer:

1.181 × 10²⁴ molecules CO₂

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

<u>Step 1: Define</u>

86.34 g CO₂

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

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Convert</u>

<u />86.34 \ g \ CO_2(\frac{1 \ mol \ CO_2}{44.01 \ g \ CO_2} )(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2} ) = 1.18141 × 10²⁴ molecules CO₂

<u>Step 4: Check</u>

<em>We are given 4 sig figs. Follow sig fig rules and round.</em>

1.18141 × 10²⁴ molecules CO₂ ≈ 1.181 × 10²⁴ molecules CO₂

4 0
3 years ago
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