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3 years ago

Are most elements on the periodic table nonmetals?

Chemistry

2 answers:

Over [174]3 years ago

8 0

Wouldn’t say most but yes

geniusboy [140]3 years ago

7 0

Answer:

THe nOme sida

Explanation:

uwba sodwpa

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Using the periodic table, choose the more reactive metal. Ag or Te

Bad White [126]

The answer is Ag.................silver

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3 years ago

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HELP ME OUT MY LOVELY CHEMISTRY CHADS I WILL GIVE YOU A KISS

vitfil [10]

Answer:

300K.

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 300L

Initial temperature (T1) = 200K

Final volume (V2) = 450L

Final temperature (T2) =..?

Since the pressure is constant, the gas is obeying Charles' law.

Using the Charles' law equation, we can obtain the new temperature of the gas as follow:

V1/T1 = V2/T2

300/200 = 450/T2

Cross multiply to express in linear form

300 x T2 = 200 x 450

Divide both side by 300

T2 = (200 x 450)/ 300

T2 = 300K

Therefore, the new temperature of the gas is 300K.

6 0

2 years ago

MATCH

almond37 [142]

I wanna say it would be 1 would be A 2 would be C 3 would be D 4 would be B and 5 would be E

hope this helps

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3 years ago

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Which sphere exists because of there is liquid iron as part of the earth's core?

Feliz [49]

The answer is the magnetosphere

4 0

3 years ago

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Consider the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g). If 0.02 moles of PCl5, 0.04 moles of PCl3, and 0.08 moles of Cl2 are combined

Furkat [3]

Answer:

The reaction quotient (Q) before the reaction is 0.32

Explanation:

Being the reaction:

aA + bB ⇔ cC + dD

$Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.

The concentration will be calculated by:

$Concentration=\frac{number of moles of solute}{Volume}$

You know the reaction:

PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).

So:

$Q=\frac{[PCl_{3} ] *[Cl_{2} ] }{[PCl_{5} ]}$

The concentrations are:

[PCl₃]= $\frac{0.04 moles}{0.5 L} =0.08 \frac{moles}{L}$

[Cl₂]= $\frac{0.08 moles}{0.5 L} =0.16 \frac{moles}{L}$

[PCl₅]= $\frac{0.02 moles}{0.5 L} =0.04 \frac{moles}{L}$

Replacing:

$Q=\frac{0.08*0.16}{0.04}$

Solving:

Q= 0.32

The reaction quotient (Q) before the reaction is 0.32

4 0

2 years ago