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svp [43]
1 year ago
10

Is using matches to light a candle physical or chemical change ?

Chemistry
2 answers:
AveGali [126]1 year ago
6 0
Chemical change because the wax particles are changing
daser333 [38]1 year ago
3 0

Answer:

It's a chemical change.

Explanation:

The process of burning (as opposed to evaporating) is a chemical reaction, a chemical change. The wax molecules are undergoing a chemical change; they are changing into different molecules by reacting with a substance in the air.

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The best answer would be A.) stay in the car.

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3 years ago
Valance shell example..​
Natali5045456 [20]

Answer:

<h2>Oxygen has six valence electrons, two in the 2s subshell and four in the 2p subshell.</h2>

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<h3>Configuration of oxygen's valence electrons as 2s²2p⁴.</h3>

Explanation:

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2 years ago
Sorry if it is blurry
Zina [86]
I really cant read it sorry i tried
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3 years ago
Read 2 more answers
Converting the following mm Hg (torr) to standard
11Alexandr11 [23.1K]

Explanation:

Given:

20.5 torr

78.6 torr

225 torr

Change into atm

Computation:

We know that;

1 torr = 0.00131579 atm

So,

1. 20.5 torr

= 20.5 x 0.00131579 atm

= 0.02697

= 0.027 atm (Approx)

2. 78.6 torr

= 78.6 x 0.00131579 atm

= 0.103090

= 0.103 atm (Approx)

3. 225 torr

= 225 x 0.00131579 atm

= 0.2960

= 0.296 atm (Approx)

5 0
3 years ago
Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a low
LenaWriter [7]

Answer:

Compound B has greater molar mass.

Explanation:

The depression in freezing point is given by ;

\Delta T_f=i\times k_f\times m..[1]

m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}

Where:

i = van't Hoff factor

k_f = Molal depression constant

m = molality of the solution

According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.

The depression in freezing point of solution with A solute: \Delta T_{f,A}

Molar mass of A = M_A

The depression in freezing point of solution with B solute: \Delta T_{f,B}

Molar mass of B = M_B

\Delta T_{f,A}>\Delta T_{f,B}

As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.

\Delta T_f\propto \frac{1}{\text{Molar mass of solute}}

M_A

This means compound B has greater molar mass than compound A,

4 0
3 years ago
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