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Savatey [412]
3 years ago
14

Anthony's homework assignment is to demonstrate that an orange has already undergone a chemical change. Which of the following

Chemistry
2 answers:
wolverine [178]3 years ago
8 0
B. Rotten orange is the correct answer. Hope this helps!
Wittaler [7]3 years ago
5 0
I agree, it’s B because you can’t change a rotten orange back into a fresh orange.
You might be interested in
How many Na and Cl molucules are in this atom? Please help
kati45 [8]
5 Na molecules and 5 Cl molecules
5 0
3 years ago
Read 2 more answers
An air sample contains 0.044% CO2? If the total pressure is 750 mm Hg. What is the partial pressure of CO2?2?
Aleksandr [31]
Partial pressure=mole fraction×Pt
x=0.044÷44(maolarmass of CO2)×Pt
x=0.044÷(44)2×Pt
x=5×10^-4×Pt
x=5×10^-4×Pt
where Pt:1atm=760mmHg
xatm=750mmHg
750×1÷760=0.99
now;5×10^-4×099=4.95×10^-4.
Pt=4.95×10^-4
8 0
3 years ago
Someone please help I will give brainliest!!!!!!!
Alex

Answer:

Option 1:

Cu(NO3)2 + Ag

hope this was helpful

7 0
3 years ago
Which element below would most likely gain 3 electrons?
Zanzabum

Answer:

Lithium

Explanation:

5 0
3 years ago
PLEASE HELP! WILL GIVE BRAINLIEST:
Ad libitum [116K]

2.45 °C

From the Ideal gas law (Combined gas law)

PV/T = P'V'/T' .....eq 1

Where:

P - initial pressure

V - initial volume

T - initial temperature

P' - final pressure

V' - final volume

T' - final temperature.

To proceed we have to make T the subject from eq.1

Which is, T = P'V'T/PV.......eq.2

We have been provided with;

Standard temperature and pressure (STP)

P = 760 mm Hg (SP - Standard Pressure)

T = 273.15 K (ST - Standard Temperature)

V = 62.65 L

P' = 612.0 mm Hg

V' = 78.31 L

T' = ? (what we require)

Therefore, we substitute the values into eq.2

T' =

.

T' = 275.60 K

T = (275.60 - 273.15) ......To °C

T = 2.45 °C

>>>>> Answer

Have a nice studies.

5 0
2 years ago
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