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oksano4ka [1.4K]
3 years ago
5

Classify each of the following as either a physical change or a chemical change:

Chemistry
2 answers:
Nady [450]3 years ago
7 0
A and B are both physical changes. The paper is still paper and the wax is still wax. C and D are both chemical changes. You are changing the log when you burn it. It is no longer a log after it is burned completely and the milk cannot be good after it has went sour. It can’t be changed back.
Taya2010 [7]3 years ago
3 0

Answer:

physical change

a b c

chemical change

d

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P4+ __02_ P4010 <br> I need help ASAP
klasskru [66]

Answer:

P_4+5O_2\rightarrow P_4O_{10}

Explanation:

Hello!

In this case, when we want to balance chemical reactions such as in this case, the idea is to equal to number of atoms of each element at each side of the equation according to the lay of conservation of mass, just as shown below:

P_4+5O_2\rightarrow P_4O_{10}

Because we have four phosphorous and ten oxygen atoms at each side.

Best regards!

8 0
3 years ago
According to the law of conservation of mass, if an element A has an atomic mass of 2 mass units and element B has an atomic mas
LenaWriter [7]
The law of conservation of mass states that mass is neither created nor destroyed. Since we have 2 g/mol of A and 3 g/mol of B then AB should be equal to the sum of their molar mass that is 

2 g/mol + 3 g/mol = 5 g/mol AB

for the case of A2B3

A2 = 2 * 2 = 4 g/mol
B3 = 3 * 3 = 9 g/mol

therefore A2B3 = 13 g/mol
5 0
4 years ago
Spell out full name of compound
Margarita [4]
Hydroxide is the full name of the compound.

Hope this Helped!

;D
Brainliest??
7 0
3 years ago
Read 2 more answers
139%
GaryK [48]

<u>Answer:</u>

The percent composition of this compound is 94%

<u>Explanation:</u>

The reaction can be formed as

2 \mathrm{Fe}+3 \mathrm{Cl}_{2} \rightarrow 2 \mathrm{FeCl}_{3}

\frac{\text { Weight of } \mathrm{Cl}_{2}}{\text { 3* Molar Mass of } \mathrm{Cl}_{2}}=\frac{\text { Weight of } \mathrm{Fe}}{2 * \text { Molar Mass of Fe }}

\frac{\text { Weight of } \mathrm{Cl}_{2}}{3 *(2 * 35.5)}=\frac{3.56}{2 * 55.8}

\text { Weight of } C l_{2}=\frac{3.56 * 3 * 71}{2 * 55.8}=6.79 \mathrm{g}

\mathrm{n}\left(\mathrm{Cl}_{2}\right)=\mathrm{m}\left(\mathrm{Cl}_{2}\right) / \mathrm{M}\left(\mathrm{Cl}_{2}\right)=6.79 / 71=0.1 \mathrm{m}

\mathrm{n}(\mathrm{Fe})=\mathrm{m}(\mathrm{Fe}) / \mathrm{M}(\mathrm{Fe})=3.56 / 55.8=0.06 \mathrm{m}

Based on no. of iron reacted,  

\mathrm{n}(\text { moles of } \mathrm{Fe})=\mathrm{n}\left(\text { moles of } \mathrm{FeCl}_{3}\right)

n = m/M

\mathrm{m}\left(\mathrm{FeCl}_{3}\right)=\mathrm{n}^{*} \mathrm{M}=0.06^{*} 162.5=9.75 \mathrm{g}

% composition ofFeCl_3  

=  (9.75 / 10.39)^{*} 100

= 94%

6 0
3 years ago
if a volume of air occupying 12.0Lat 20celsius is heated to a new temperature of 100 celsius,what would be the new volume
emmasim [6.3K]

Answer:

V₂ = 15.3

Explanation:

Given data:

Initial volume = 12.0 L

Initial temperature = 20°C

Final temperature =100°C

Final volume = ?

Solution:

First of all we will convert the temperature into kelvin.

20°C + 273 = 293 K

100°C + 273 = 373 K

Formula:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 12.0 L × 373 K / 293 k

V₂ = 4476 L.K /293 k

V₂ = 15.3

V₂ = 1566 L.K / 298 K

V₂ = 5.3 L

6 0
3 years ago
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