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2 years ago

Objects in free-fall ___.

Chemistry

2 answers:

Ira Lisetskai [31]2 years ago

6 0

Answer:

Explanation:

mote1985 [20]2 years ago

3 0

C) Are accelerating faster than 9.8 m/s/s

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How much energy is released when 0.40 mol C6H6(g) completely reacts with oxygen.

Mashutka [201]

You have to find the gram formula mass of C6H6 then do mass (g) = mol x GFM

5 0

2 years ago

Will bromine react with sodium and why?

Archy [21]

When you write down the electronic configuration of bromine and sodium, you get this

Na:
Br:

So here we the know the valence electrons for each;

Na: (2e)
Br: (7e, you don't count for the d orbitals)

Then, once you know this, you can deduce how many bonds each can do and you discover that bromine can do one bond since he has one electron missing in his p orbital, but that weirdly, since the s orbital of sodium is full and thus, should not make any bond.

However, it is possible for sodium to come in an excited state in wich he will have sent one of its electrons on an higher shell to have this valence configuration:

where here now it has two lonely valence electrons, one on the s and the other on the p, so that it can do a total of two bonds.That's why bromine and sodium can form

4 0

3 years ago

Help me please ❤❤❤

musickatia [10]

Answ burh i dont even know

Explanation: i dont even know this

7 0

2 years ago

What substance is reduced in the reaction 2Fe 2+ + CI2 = 2Fe 3+ + 2CI- ?

atroni [7]

Answer:

$Cl_{2}$ is reduced in the reaction

Explanation:

The given reaction is

$2Fe^{2+}+Cl_{2} \to 2Fe^{3+}+2Cl^{-}$

The oxidation number of $Fe$ is changed from $+2\, \to \, +3$

$Fe^{2+} \to Fe^{3+}+e^{-}$

And The oxidation number of $Cl$ is changed from $0\, \to \, -1$

$Cl_{2}^{0} + 2e^{-} \to 2Cl^{-}$

Hence, $Fe^{2+}$ is oxidized and $Cl_{2}$ is reduced

5 0

3 years ago

Assume the hydrolysis of ATP proceeds with ΔG′° = –30 kJ/mol. ATP + H2O → ADP + Pi Which expression gives the ratio of ADP to AT

Andru [333]

Answer:

$6.14\cdot 10^{-6}$

Explanation:

Firstly, write the expression for the equilibrium constant of this reaction:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

$\Delta G^o = -RT ln K_{eq}$

From here, rearrange the equation to solve for K:

$K_{eq} = e^{-\frac{\Delta G^o}{RT}}$

Now we know from the initial equation that:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Let's express the ratio of ADP to ATP:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}$

Substitute the expression for K:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}$

Now we may use the values given to solve:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}$

7 0

3 years ago