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2 years ago

GIVING BRAINLIEST AND THE REST OF MY POINTS!!!!! :D

Chemistry

1 answer:

Alik [6]2 years ago

6 0

It’s d I did this before

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Into which molecule are all the carbon atoms in glucose ultimately incorporated during cellular respiration?

pav-90 [236]

Answer:

The correct option is: carbon dioxide (CO₂)

Explanation:

Cellular respiration refers to the process of generation of energy in the form of adenosine triphosphate (ATP) from various nutrients such as carbohydrates, fats, and proteins.

Aerobic cellular respiration involves the oxidation of nutrients such as glucose (C₆H₁₂O₆), by molecular oxygen (O₂) to give carbon dioxide (CO₂), water (H₂O) and heat energy.

Reaction involved: (glucose) C₆H₁₂O₆ (s) + 6 O₂ (g) → 6 CO₂ (g) + 6 H₂O (l) + heat energy

Therefore, in a cellular respiration, all the carbon atoms of the glucose molecules, form carbon dioxide.

3 0

2 years ago

What is the formal charge in SeCl4?

Nuetrik [128]

The answer is 0 !!!!!!!

3 0

1 year ago

Pentaborane-9, B5H9, is a colorless, highly reactive liquid that will burst into flame when exposed to oxygen. The reaction is 2

mina [271]

Answer: The amount of energy released per gram of $B_5H_9$ is -71.92 kJ

Explanation:

For the given chemical reaction:

$2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]$

Taking the standard enthalpy of formation:

$\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(5\times (1271.94))+(9\times (-285.83))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H^o_{rxn}=-9078.57kJ$

We know that:

Molar mass of pentaborane -9 = 63.12 g/mol

By Stoichiometry of the reaction:

If 2 moles of $B_5H_9$ produces -9078.57 kJ of energy.

Or,

If $(2\times 63.12)g$ of $B_5H_9$ produces -9078.57 kJ of energy

Then, 1 gram of $B_5H_9$ will produce = $\frac{-9078.57kJ}{(2\times 63.12)}\times 1g=-71.92kJ$ of energy.

Hence, the amount of energy released per gram of $B_5H_9$ is -71.92 kJ

8 0

2 years ago

The oxidation of ammonia produces nitrogen and water via the following reaction: 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l) Suppose the

Sonbull [250]

Answer:

The rate of consumption of $NH_{3}$ is 2.0 mol/L.s

Explanation:

Applying law of mass action to this reaction-

$-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=-\frac{1}{3}\frac{\Delta [O_{2}]}{\Delta t}=\frac{1}{2}\frac{\Delta [N_{2}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}$

where $-\frac{\Delta [NH_{3}]}{\Delta t}$ represents rate of consumption of $NH_{3}$ , $-\frac{\Delta [O_{2}]}{\Delta t}$ represents rate of consumption of $O_{2}$ , $\frac{\Delta [N_{2}]}{\Delta t}$ represents rate of formation of $N_{2}$ and $\frac{\Delta [H_{2}O]}{\Delta t}$ represents rate of formation of $H_{2}O$ .