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2 years ago

You are designing a new product that requires the use of a stable, nonreactive

Chemistry

2 answers:

klasskru [66]2 years ago

3 0

Oxygen is a no reactive gas

ratelena [41]2 years ago

3 0

Answer:

Argon

Explanation:

just did test and people were confused in comments

hope this helps

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What mass of powdered drink mix is needed to make a 0.5 M solution of 100 mL?

Maru [420]

Answer:

There is 17,114825 g of powdered drink mix needed

Explanation:

<u>Step 1 :</u> Calculate moles

As given, the concentration of the drink is 0.5 M, this means 0.5 mol / L

Since the volume is 100mL, we have to convert the concentration,

⇒0.5 / 1 = x /0.1 ⇒ 0.5* 0.1 = x = 0.05 M

This means there is 0.05 mol per 100mL

<u>Step 2 </u>: calculate mass of the powdered drink

here we use the formula n (mole) = m(mass) / M (Molar mass)

⇒ since powdered drink mix is usually made of sucrose (C12H22O11) and has a molar mass of 342.2965 g/mol.

0.05 mol = mass / 342.2965 g/mol

To find the mass, we isolate it ⇒0.05 mol * 342.2965 g/mol = 17,114825g

There is 17,114825 g of powdered drink mix needed

3 0

2 years ago

Read 2 more answers

A soft drink contains 11.5% sucrose (C12H22O11) by mass. How much sucrose, in grams, is contained in 355 mL (12 oz) of the soft

luda_lava [24]

Answer:

42.5 g

Explanation:

Calculate the mass of the soft drink given the density and volume:

355 mL × 1.04 g/mL = 369.2 g

Now calculate the mass of sucrose given the percentage:

0.115 × 369.2 g = 42.46 g

Rounded to 3 significant figures, the mass is 42.5 g.

5 0

2 years ago

Read 2 more answers

A sample of an unknown compound with a mass of 0.847 g has the following composition: 50.51 % fluorine and 49.49 % iron. When th

sergejj [24]

Answer: 0,4278g of F and 0,4191g of Fe

Explanation: it's possible to calculate the mass of each element by multiplying the percentage (decimal) of the element by the mass of the compound.

For Fluorine (F)

0,847g * 0,5051 = 0,4278g of F

For iron (Fe)

0,847 * 0,4949 = 0,4191g of Fe

This is determined because even when the compound is decomposed, due to conservative law of mass, the decomposition process do not affect the amount of matter, so the mass of the elements remain even if they are separated from the original molecule.

At the end, the sum of the elements masses should be the total mass of the compound.

4 0

2 years ago

Read 2 more answers

Hydrogen sulfide (H2S) is a common and troublesome pollutant in industrial wastewaters. One way to remove H2S is to treat the wa

Kryger [21]

Explanation:

As the given reaction is as follows.

$H_{2}S(aq) + Cl_{2}(aq) \rightarrow S(s) + 2H^{+}(aq) + 2Cl^{-}(aq)$

So, according to the balanced equation, it can be seen that rate of formation of $Cl^{-}$ will be twice the rate of disappearance of $H_{2}S$ .

And, it is known that rate of disappearance of reactant will be negative and rate of formation of products will be positive value.

This means that,

Rate of the reaction = -Rate of disappearance of $H_{2}S$

= $k[H_{2}S][Cl_{2}]$

= $(3.5 \times 10^{-2}) \times (2 \times 10^{-4}) \times (2.8 x 10^{-2})$

= $1.96 \times 10^{-7}$ M/s

Therefore, calculate the rate of formation of $Cl^{-}$ as follows.

Rate of formation of $Cl^{-}$ = $2 \times 1.96 \times 10^{-7}$

= $3.92 \times 10^{-7}$ M/s

Thus, we can conclude that the rate of formation of $Cl^{-}$ is $3.92 \times 10^{-7}$ M/s.

5 0

2 years ago

When 24 mol of methanol and 15 mol of oxygen combine in the combustion reaction, 2 ch3oh(

umka21 [38]

The balanced equation for combustion is as follows;
2CH₃OH + 3O₂ ---> 2CO₂ + 4H₂O
The stoichiometry of CH₃OH to O₂ is 2:3
the limiting reagent is the reactant that is fully consumed during the reaction. The amount of product formed is directly proportional to the amount of limiting reactant produced. The excess reagent is the reactant that is provided in excess and is not fully used up, there will be an amount of this reagent remaining after the reaction.
If methanol is the limiting reactant,
If 2 mol of methanol reacts with 3 moles of O₂
Then 24 mol of methanol reacts with - 3/2 x 24 = 36 mol of O₂ should be present
But only 15 mol of O₂ is present, therefore O₂ is the limiting reactant and methanol is in excess.
3 mol of O₂ reacts with 2 mol of CH₃OH
then 15 mol of O₂ reacts with 2/3 x 15 = 10 mol of CH₃OH

Excess reactant is methanol, 10 mol are used up therefore 24 - 10 mol = 14 mol are remaining at the end of the reaction

8 0

2 years ago