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MariettaO [177]
2 years ago
8

You are given a piece of paper and a match. The paper has a mass of 2.5 g. You then light the match and light the piece of paper

on fire. After it burns, the remaining bits of paper weigh 0.5 g. Does this demonstration violate the conservation of mass? Explain why or why not?
Chemistry
1 answer:
kow [346]2 years ago
5 0

Answer:

No

Explanation:

<em>No. </em>T<em>he demonstration does not violate the conservation of mass.</em>

<u>The law of conservation of mass states that mass can neither be created nor destroyed in a reaction. However, mass can be converted from one form to another during the reaction.</u>

In this case, even though the remaining bits of paper weigh 0.5 g while the original paper weighed 2.5 g, the ashes and smoke/gas from the burning will all add up to the lost weight of the paper.

<em>The burned part has been converted into other forms. If the smoke/gas and the ashes are properly captured, they will mark up with the weight of the remaining paper to give the weight of the original paper. </em>

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In science class, Blaine’s teacher puts one glow stick in a cup of hot water and another glow stick in a cup of cold water. She
timofeeve [1]

Answer:

The glow stick in hot water will be brighter

Explanation:

The glow stick in hot water will be brighter than the glow stick in cold water because the heat from the hot water will cause the molecules in the glow stick to move faster. The faster the molecules move in the glow stick, the sooner and brighter the reaction will be. The cold water will cause molecules to move slowly and it will take longer for the reaction to occur, which will also make it less bright.

3 0
2 years ago
Read 2 more answers
A 55.0 L steel tank at 20.0 ∘C contains acetylene gas, C2H2, at a pressure of 1.39 atm. Assuming ideal behavior, how many grams
JulijaS [17]

Answer:

PV=nRT

n = PV/RT

n = m/Mm

m/Mm = PV/RT

m = MmPV/RT

T in kelvin = T Celsius + 273.15 = 293.15 K

m = (26.04 x 1.39 x 55)/(0.08206 x 293.15)

mass in grams = 82.8 grams  

Explanation:

Ideal gases formula is PV=nRT, where:

P is the pressure (1.39 atm in this case)

V is the volume (55.0 L in this case)

R is the gas constant (0.08206 L.atm/K.mole)

T is the temperature (20.0C) should be converted to Kelvin

all the unit should correspond to the one in the R.

we also know that to find the mass, we can use number mole with the formula number of mole(n) = mass (m) divided by the molar mass (Mm). therefore we substituted that in the formula and make (m) the subject of the formula.

we found the mass to be 82.8 grams

7 0
2 years ago
The reaction of benzene with (CH3)3CCH2Cl in the presence of anhydrous aluminum chloride produces principally which of these?
Sedaia [141]

Answer:

See explanation and image attached

Explanation:

When the carbocation is formed by the action of AlCl3 on the (CH3)3CCH2Cl, a primary carbocation is formed. The formation of the carbonation is followed by a 1,2-alkyl shift to give a tertiary carbocation which subsequently adds to the benzene ring as shown in the image attached to this answer.

6 0
2 years ago
A block of aluminum weighing 140 g is cooled from 98.4°C to 62.2°C with the release of 1080 joules of heat. From this data, calc
GrogVix [38]

<u>Answer:</u>

Specific heat of a substance is the value that describe how the added heat energy of substance has the impact on its temperature.

Unit is <em>(\frac {J}{Kg.K})</em>

<em>C = Q/m. ∆T</em>

<em>C – Specific heat (\frac {J}{Kg.K})</em>

<em>Q- heat energy (J)</em>

<em>M – Mass (Kg)</em>

<em>∆T- change in temperature (K) </em>

<u>Explanation:</u>

<em>Given data:</em>

<em>M= 140 g = 0.14 Kg</em>

<em>Q – 1080 Joules.</em>

<em>∆T – 98.4 – 62.2 = 36.2</em>

Substituting  the given data in Equation

<em>Specific heat of Aluminium  = \frac {1080}{(0.14 \times 36.2)} = 213.10 (\frac {J}{Kg.K})</em>

3 0
2 years ago
Read 2 more answers
How many atoms are centered on the (100) plane for the fcc crystal structure?
Setler [38]

Here, we have to get the number of atoms present in the 100 plane of the FCC crystal lattice.

There will be 2 atoms in 100 plane of FCC crystal lattice.

In the face centered crystal (FCC) lattice there are atoms at each corner of the cube and each are shared by 4 another atoms. And an atom is present at the face of the crystal.

For the 100 plane of the Miller indices the intercepts are a, ∞, ∞ or 2a, ∞, ∞.

Thus, for the 4 atoms of the corner at the cube shared by 4 other atoms will contribute, 4 × \frac{1}{4} = 1 and the un-shared atoms at the face will contribute another 1, which make the total atom 1 + 1 = 2.

5 0
2 years ago
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