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saw5 [17]
3 years ago
12

Why should we put a glass on candles for experiments and why should we put candles on sand trays ?​

Chemistry
1 answer:
Lina20 [59]3 years ago
3 0

Answer:

1, you should use candles for experiments because, You created an area of low pressure! When the experiment is run you can see tiny bubbles escaping under the glass which shows that the increased air pressure from the heated air as the candle burns. Once the candle runs out of oxygen, the candle burns out and the remaining air inside cools down.

2. you should put candles on sand trays, Pour layers of sand into a glass vase, alternating colors with each layer. For a more interesting look, try uneven layers in various heights. Keep layering sand until you're a few inches from the top of the vase. Place your votive candle in the center of the sand.

Explanation:

its scientific proven

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A 2.0 L flexible container holds 5.0 moles of oxygen (O2) gas. An additional 15.0 moles of nitrogen gas (N2) is added to the con
frozen [14]
Answer 8.0 L.

2.0L / 5.0 moles = x / 20.0 => x = 20 / 5 * 2 = 8
4 0
3 years ago
Read 2 more answers
Chemical formula for tin iv carbide
aniked [119]

Answer:

Explanation:

Sn(WC)2

if it is tungsten carbide this should be correct but there are many versions of carbide

Sn(MC2)2

could also be possible

the 2 next to MC should be a subscript

3 0
3 years ago
Describe how visible spectroscopy could be used to determine the order of a reaction in which a single colorless reactant produc
d1i1m1o1n [39]

Answer:

Explanation:

In a reaction, where, one of the reactant produces a colored product, visible spectroscopy can be used to determined the order of a reaction, the change in concentration of the reactant which forms the colored product is determined by absorbance measurement over time. The data for the concentration and time are plotted on the y and x axis and If we get a straight line it is a zero-order reaction. If instead, a plot of ln[concentration] versus time gives a straight line, it is a first order reaction. However, If 1/concentration versus time gives a straight line, it is a second order reaction kinetics. The other reactants may be changed while keeping this reactant as constant and change on rate of the reaction is observed to see If the other reactant affects the reaction or not.

8 0
4 years ago
When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 27.6 g of carbon were burned in the presence of
In-s [12.5K]

Answer:

The answer to your question is: 101.2 g of CO2

Explanation:

C = 27.6 g

O₂ = 86.5 g   remained 12.9 g

O₂ that reacted = 86.5 - 12.9 = 73.6 g

                     C     + O₂      ⇒        CO₂      The equation is balanced

                    27.6    73.6                 ?

MW               12        32                  44

Rule of three

                        12 g of C------------------  44 g  CO2

                       27.6 g C  ------------------    x

                  x = 27.6(44)/12 = 101.2 g of CO2

                       32 g of O2 ---------------    44 g of CO2

                        73.6 g of O2 ------------      x

                  x = 73.6(44)/32 = 101.2 g of CO2

6 0
3 years ago
What mass (g) of magnesium nitride (Mg3N2) can be made from the reaction of 1.22 g of magnesium with excess nitrogen? __Mg + __N
Citrus2011 [14]
<h3>Answer:</h3>

1.69 g Mg₃N₂

<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 Table

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

[RxN - Unbalanced] Mg + N₂ → Mg₃N₂

[RxN - Balanced] 3Mg + N₂ → Mg₃N₂

[Given] 1.22 g Mg

[Solve] grams Mg₃N₂

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

[RxN] 3 mol Mg → Mg₃N₂

[PT] Molar Mass of Mg - 24.31 g/mol

[PT] Molar Mass of N - 14.01 g/mol

Molar Mass of Mg₃N₂ - 3(24.31) + 2(14.01) = 100.95 g/mol

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                      \displaystyle 1.22 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg})(\frac{1 \ mol \ Mg_3N_2}{3 \ mol \ Mg})(\frac{100.95 \ g \ Mg_3N_2}{1 \ mol\ Mg_3N_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 1.68873 \ g \ Mg_3N_2

<u>Step 4: Check</u>

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

1.68873 g Mg₃N₂ ≈ 1.69 g Mg₃N₂

8 0
3 years ago
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