Given the standard enthalpies of formation for the following substances, determine

Can someone please help? We're studying astronomy and I have no idea what to name it. I was never good with names. ‍♀️

almond37 [142]

I do not understand your question, what is the question, name what??

5 0

4 years ago

How many moles of iron are there in 55.85g of Fe3O4

soldier1979 [14.2K]

Answer:

• Molecular mass of Iron (III) tetraoxide

$\dashrightarrow \: { \tt{(56 \times 3) + (16 \times 4)}} \\ = { \tt{168 + 64}} \\ = { \tt{232\:g}}$

[ molar masses: Fe → 56, O → 16 ]

$\dashrightarrow \:{ \rm{232 \: g \: = 1 \: mole}} \\ \\ \dashrightarrow \: { \rm{55.85 \: g = ( \frac{55.85}{232}) \: moles }} \\ \\ \dashrightarrow \: { \boxed{ \tt{ = 0.24 \: moles}}}$

8 0

3 years ago

For the reaction: 3 H2(g) + N2(g) <--> 2 NH3(g), the concentrations at equilbrium were [H2] = 0.10 M, [N2] = 0.10 M, and [

masya89 [10]

The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵

<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>

It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.

Hence, the equilibrium constant of the reaction in discuss is;

K = [5.6]²/[0.10]³[0.10]

k = 5.6² × 10⁴

k = 3.136 × 10⁵

K = 3.1 × 10⁵.

Read more on equilibrium constant;

brainly.com/question/1619133

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5 0

2 years ago

Which of the following choices has the ecological levels listed from largest to smallest?

wariber [46]

Answer:

the answer is something

Explanation:

3 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

4 years ago