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Veronika [31]
3 years ago
6

Check all examples that would a chemical property.

Chemistry
1 answer:
fenix001 [56]3 years ago
3 0

Answer:

d. reactivity to acid

f. reacts to air

Are chemical properties

Explanation:

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i have an unknown volume of gas at a pressure of 0.50 atm a temperature of 325 k. If i raise the pressure to 1.2 atm, decrease t
Ronch [10]
Assuming that the number of mols are constant for both conditions:
\frac{P_1V_1}{T_1} =  \frac{P_2V_2}{T_2}
Now you plug in the given values. V_1 is the unknown. 
\frac{0.50 atm*V_1}{325K} = \frac{1.2 atm* 48 L}{230K}

Separate V_1
V_1= \frac{1.2 atm* 48 L * 325K }{230K*0.50 atm }
V= 162.782608696 L 

There are 2 sig figs

V= 160 L
3 0
3 years ago
Magnesium and calcium are in the same group on the periodic table.
SOVA2 [1]

Answer: They both have two valence electrons.

Explanation:

Atoms with the same number of valence electrons normally have similar properties.

8 0
2 years ago
Acrostic words or phrases that describe street dance​
Basile [38]
Poppin? dance??? LOL
6 0
2 years ago
When an excited electron in a hydrogen atom falls from ????=5 to ????=2, a photon of blue light is emitted. If an excited electr
poizon [28]

Answer:

n = 3

Explanation:

Given the formula for the transition energy of an atom with 1 electron:

E=-13.6*Z^{2}*(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}} ) eV

For the H transition n=5 to n=2:

E=-13.6*(\frac{1}{4}-\frac{1}{25} ) eV=-2.856 eV

Then we solve for nf with Z=2 (Helium)

n_{f}=\sqrt{\frac{n_{i}^{2}*Z^{2}*13.6 eV }{2.856eV*n_{i}^{2}+13.6eV*Z^{2}} }

n_{f}=\sqrt{\frac{4^{2}*2^{2}*13.6 eV }{2.856eV*4^{2}+13.6eV*2^{2}} }=3

Is near 3, actually the energy of the transitions are:

H (5⇒2) = -2.85 eV = 434 nm (Dark blue)

He (4⇒3) = -2.64 eV = 469 nm (Light blue)

I thought it was cool to see the actual colors. Included them.

4 0
3 years ago
Cooking Oil
Grace [21]

Answer:

corn oil ,2.50 will be receiving

4 0
3 years ago
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