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Lera25 [3.4K]
4 years ago
7

If one molecule of glucose (C6H12O6) reacts completely with three molecules of oxygen (O2), what is the TOTAL number of atoms th

at will be in the products? A) 24 B) 30 C) 38 D) 48
Chemistry
2 answers:
Dmitriy789 [7]4 years ago
8 0
I believe it would be 30 because there are 24 atoms in one molecule of glucose and 6 atoms in three molecules of oxygen.
USPshnik [31]4 years ago
7 0

Answer;

30 atoms

Explanation;

When glucose, C6H12O6, reacts with oxygen it produces carbon dioxide and water. If a molecule of glucose reacts completely with three molecules of oxygen, then 30 atoms would be produced because there are 24 atoms in glucose and 6 atoms in oxygen molecules.

The equation for the reaction would be;

C6H12O6 + 6O2(g) → 6CO2(g) + 6 H2O(l)

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6/10
hichkok12 [17]

Answer: B

Explanation: I answered randomly

5 0
3 years ago
What is the meaning of the following metric prefixes kilo,deco,centi,milli,micro and nono
BigorU [14]
Kilo: a prefix used on metric units to indicate a multiple of 1000. For example 1kg = 1000 g
4 0
4 years ago
Given that ΔH = −571.6 kJ/mol for the reaction 2 H2(g) + O2(g) → 2 H2O(l), calculate ΔH for these reactions. (a) 2 H2O(l) → 2 H2
pashok25 [27]

Answer : The value of \Delta H for the reaction is +571.6 kJ/mole.

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction is,

2H_2(g)+O_2(g)\rightarrow 2H_2O(l)     \Delta H_1=-571.6kJ/mole

Now we have to determine the value of \Delta H for the following reaction i.e,

2H_2O(l)\rightarrow 2H_2(g)+O_2(g)    \Delta H_2=?

According to the Hess’s law, if we reverse the reaction then the sign of \Delta H change.

So, the value \Delta H_2 for the reaction will be:

\Delta H_2=-(-571.6kJ/mole)

\Delta H_2=+571.6kJ/mole

Hence, the value of \Delta H for the reaction is +571.6 kJ/mole.

6 0
3 years ago
Suppose that 0.00150 moles of CO2 (44.0 g/mol) effuse out of a pinhole in 1.00 hour. How many moles of N2 (28.0 g/mol) would eff
kenny6666 [7]

Answer:

I was having trouble with this question idk the answer

Explanation:

4 0
3 years ago
Read 2 more answers
I don't quite understand it
irina1246 [14]
So, you need to have same ammount of atoms on the left and on the right side of the equation. You need to count the ammount of attoms of every substance on the left, and make sure that on the right side the ammount is same. For example in the 1st one it’s 6Sn+2P4=2Sn3P4, so that you have 6atoms of Sn on the left and 6 atoms of Sn on the right, same with the P
6 0
3 years ago
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