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3 years ago

This question is about metal oxides. When sodium is heated in oxygen, sodium oxide is produced.

Chemistry

1 answer:

vodomira [7]3 years ago

5 0

Answer:

This is very easy Cuz We have 2Na2O We have O2 so thats molecule of Oxygen and its same on product We need to balance Na on start We have 1 on product We have 2 so Just put 2 at start....

Explanation:

Sorry for bad english not my first language :(

2Na+O2-->2Na20

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You have three drinks in front of you. you know how they are made, but are unsure which one has the highest concentration of jui

GuDViN [60]

(I think) you would have to divided the number i dont know the exact answer but ya hope this help

4 0

3 years ago

Read 2 more answers

In E. coli, the enzyme hexokinase catalyzes the reaction: Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Ke

kondor19780726 [428]

Answer:

Explanation:

Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.

In the living E. coli cells,

[ATP] = 7.9 mM;

[ADP] = 1.04 mM,

[glucose] = 2 mM,

[glucose 6-phosphate] = 1 mM.

Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.

The reaction is given as

Glucose + ATP → glucose 6-phosphate + ADP

Now reaction quotient for given equation above is

$q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}$

$q=\frac{(1mm)\times (1.04 mm)}{(7.9mm)\times (2mm)} \\\\=6.582\times 10^{-2}$

so,

$q$ ⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable until q = Keq

4 0

3 years ago

During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce

swat32

Answer: 16.32 g of $O_2$ as excess reagent are left.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol$

$\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol$

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

According to stoichiometry :

2 moles of $SO_2$ require = 1 mole of $O_2$

Thus 0.34 moles of $SO_2$ will require= $\frac{1}{2}\times 0.34=0.17moles$ of $O_2$

Thus $SO_2$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

Moles of $O_2$ left = (0.68-0.17) mol = 0.51 mol

Mass of $O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g$

Thus 16.32 g of $O_2$ as excess reagent are left.

3 0

2 years ago

A plot of binding energy per nucleon Eb A versus the mass number (A) shows that nuclei with a small mass number have a small bin

Illusion [34]

Answer:

6He = 4.90 MeV/Nucleon

8Li = 5.18 MeV / Nucleon

62Ni = 8.82 MeV / Nucleon

115 In = 8.54 MeV / Nucleon

Explanation:

Our strategy here is to remember that when the mass of a given nuclei is calculated from the sum of the mass of its protons and neutrons, this mass is greater than the actual value . This is the mass defect.

Now this mass defect we can convert to energy by utilizing Einstein´s equation, E = mc².This is the binding energy.

For 6He with actual mass 6.0189 u ( He has Z = 2, that is 2 protons )

mass protons = 2 x 1.0078 u = 2.0516 u

mass neutrons = 4 x 1.0087 u = 4.0348 u

predicted mass = (2.0516 + 4.0348) u = 6.0504 u

mass defect = (6.0504 - 6.0189) u = 0.0315 u

Now we need to convert this mass expressed in atomic mass units to kilograms ( 1 u = 1.66054 x 10⁻²⁷ Kg )

0.0315 u x 1.66054 x 10⁻²⁷ Kg =5.231 x 10⁻²⁹ Kg

E =5.231x 10⁻²⁸ Kg x (3 x 10⁸ m/s )² = 4.707 x 10⁻¹²J

Finally we will convert this energy in Joules to eV

E = 4.707 x 10⁻¹² J x 6.242 x 10¹⁸ eV/J = 2.94 x 10⁷ eV = 29.4 MeV

E per nucleon for 6He = 29.4 MeV / 6 =4.90 MeV / Nucleon

Now the calculations for the rest of the nuclei are performed in similar manner with the following results:

8Li = 5.18 MeV / Nucleon

62Ni = 8.82 MeV / Nucleon

115 In = 8.54 MeV / Nucleon

8 0

3 years ago

How many moles of bromine atoms are in 2.60×10^2 grams of bromine?

AlladinOne [14]

Your answer is 3.25 moles of Bromine

4 0

3 years ago