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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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Sodium borohydride, NaBH4, and boron trifluoride, BF3, are compounds of boron. What are the shapes around boron in the borohydri
STatiana [176]
Compound 1: Sodium borohydride
In sodium borohydride (NaBH4), B is a central metal.
Electronic configuration of B is 1s2 2s2 2p1.
B undergoes sp3 hybridization in NaBH4, to generate 4 hybrid orbitals. These hybrid orbitals, forms sigma bond with 4 'H' atoms. Due to this, the structure of sodium borohydride in tetrahedral.
........................................................................................................................

Compound 2: B<span>oron trifluoride
</span>In boron trifluoride (BF3), B is a central metal.
Electronic configuration of B is 1s2 2s2 2p1.
B undergoes sp2 hybridization in NaBH4, to generate 3 hybrid orbitals. These hybrid orbitals, forms sigma bond with 3 'H' atoms. Due to this, the structure of <span>boron trifluoride</span> is <span>triangular planner</span>.
7 0
3 years ago
Read 2 more answers
By how much will the ph change if 0.025 mol of hcl is added to 1.00 l of the buffer that contains 0.15 m hc2h3o2 and 0.25 m c2h3
Setler [38]
According to the reaction equation:

              CH3COO-  + H+   → CH3COOH 

initial         0.25                             0.15

change     - 0.025                          + 0.025

Equ         (0.25-0.025)                 (0.15 + 0.025)

first, we have to get moles acetate and moles acetic acid:

moles of acetate = 0.25 - 0.025 = 0.225 moles

∴ [CH3COO-] = 0.225 mol / 1 L = 0.225 M

moles of acetic acid = 0.15 + 0.025 = 0.175 moles

∴ [ CH3COOH] = 0.175 mol / 1L = 0.175 M 

Pka = -㏒ Ka 

       = -㏒ 1.8 x 10^-5
       
       = 4.74

from H-H equation we can get the PH value:

PH = Pka + ㏒ [acetate / acetic acid]

PH = 4.74 + ㏒[0.225/0.175]

∴ PH = 4.8
3 0
3 years ago
All BUT one is a quantitative observation. 420 dogs .667 cm big grasshoppers
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Big grasshoppers, because it does not involve a number.

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7 0
3 years ago
Evaluate each statement and determine which statement is not a determining factor in the formation of intrusive igneous rocks? G
sladkih [1.3K]

Answer: The statement that is not a determining factor in formation of intrusive igneous rocks is 'None of the above' and Magma cools very fast beneath the Earth's surface.

Explanation:

Rocks are naturally occurring solid materials that are made up of different types of minerals which affects it's texture and colours. The three main types of rocks are:

--> sedimentary rocks,

--> metamorphic rocks and

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Igneous rocks are formed from the crystallization and solidification of hot molten rocks which originates from deep within the earth. Depending on where the molten rock solidifies, the igneous rock is divided into two, namely:

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--> extrusive igneous rock.

The INTRUSIVE IGNEOUS rocks are formed within or deep beneath the earth surface when the hot magma SLOWLY cools over millions of years until it solidifies. The slow cooling allows large crystals to grow.

Therefore the statements (None of the above and Magma cools very fast beneath the Earth's surface) are not a determining factor.

7 0
3 years ago
How do you do this ???
trapecia [35]
1. 254 cal = 1,062.736 joules
2. 126 cal = 527.184 joules 
3. 98 cal = 410.032 joules
4. 704 cal = 2,945.536 joules
5. 682 cal = 2,853.488 joules
5 0
3 years ago
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