N writing a chemical equation that produces hydrogen gas, the correct representation of hydrogen gas is

An over-the-counter medicated foot powder contains camphor, eucalyptus oil, lemonoil, and zinc oxide. Only one of the ingredient

attashe74 [19]

Answer:

Zinc oxide is the antimicrobial in the medicated foot powder.

Explanation:

Zinc oxide (ZnO) is used to treat several skin conditions such as acne, dermatitis, eczema, etc because it has antibacterial and antifungal properties. This property of the ZnO depends on its size i.e., the smaller the size of ZnO, the stronger antimicrobial activity it has. So, the ZnO nanoparticles are more efficient.

The zinc oxide interacts with the sulfur atoms in microbial proteins and denatures them and destroys their function. This action completely inhibits microbial growth. The treatment with zinc oxide causes the reduced production of conidia in fungi, damages their hyphae and inhibits their ability to produce mycotoxins.

The compounds containing zinc such as ZnSO4, Zn(ClO4)2, etc also have antifungal and antimycotoxin properties that can cause changes in the fungi cell structure. Also, the free radicals formed on the surface of the ZnO nanoparticles can cause damage to the lipids in the bacterial cell membranes that can lead to the leakage and breakdown of the bacterial cell membranes.

5 0

2 years ago

What changes could be made to optimize the manufacturing process

Blizzard [7]

Manufacturers can generate new value minimize cost and increase operational stability by focusing on 4 broad areas management Supply Circle product design and value recovery

7 0

3 years ago

A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

7 0

2 years ago

If 8.6 g of ch4 and 5.9 g of o2 react, what is the mass, in grams, of h2o that is produced?

Alenkasestr [34]

This may seem confusing because they give you two masses, but all you have to do is pick one to do the calculations. Personally, I would pick O2, since the molar mass is easier to calculate. The answer would be 3.3 g (rounded for sig figs). To get this, first take the 5.9 grams of O2 and convert it to moles by dividing by the molar mass of oxygen gas, which is 32. Then, multiply both by the mole-mole ratio, which is 2:2, or simply 1:1. After that, multiply that by 18g, which is the molar mass of water to get grams of water.

REMEMBER, you have to write and balance the chemical equation before you can do any of that work.
That happens to be CH4 + 2O2 => CO2 + 2H2O

7 0

2 years ago

They all have the same choices but one of them its not gonna be used.

Dima020 [189]

Answer:

limiting reactant

Explanation:

it is consumed totally

3 0

2 years ago