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AveGali [126]
3 years ago
13

Help me, ANSWER ALL 6 QUESTION, if you completed, you will have 30 POINTS

Chemistry
1 answer:
kondaur [170]3 years ago
5 0

Answer:

ONE: spoon/ catapult

TWO: ball holder

THREE: the joint

FOUR: the rotater

FIVE: I think it is a lever that picks up the big spoon/ catapult

SIX: I don't know the chapter

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Where on the table would a group's<br> ionization energy be the greatest?
Ymorist [56]

Answer:

upper left

Explanation:

it is a trend on the periodic table. Ionization energy increases from left to right(->) I t also increases down to up

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3 years ago
Can someone give me an example of balancing equations with a solution that is simple?​
Monica [59]

Explanation:

H _{2}O _{2(aq)} →H _{2}O _{(l)}  + O _{2}(g) \\ solution  : 2 \:  and\: 2

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3 years ago
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He (Helium) is _________ atomic molecule.
babunello [35]

Answer:

Monatomic molecule

Explanation:

Each helium atom has 2 electrons, which is already the maximum no. of electrons that can fit in the first electron shell. When the outermost electron shell is full (2 for the first layer, 8 for others), the atom is stable.

Helium atom itself is already stable, so it doesn't need to combine with another helium atom to form a molecule, hence we call it monatomic.

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which of the following experiments follow ethical standards? A.)performing test on animals when the information can be obtained
olasank [31]
The answer is D because it sounds like the most ethical answer. The other choices are wrong because the could possibly hurt someone or an animal.
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3 years ago
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For the reaction C2H4(g) + H2O(g) --&gt; CH3CH2OH(g)
Dominik [7]

Answer : The value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

Explanation :

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

where,

\Delta G^o = standard Gibbs free energy  = ?

\Delta H^o = standard enthalpy = -45.6 kJ = -45600 J

\Delta S^o = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)

\Delta G^o=-12666.6J=-12.7kJ

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln k

where,

\Delta G^o = standard Gibbs free energy  = -12666.6 J

R = gas constant  = 8.314 J/K.mol

T = temperature  = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k

k=3.35\times 10^{2}

Therefore, the value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

3 0
3 years ago
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