How is plastic made?

Magnesium reacts with a certain element to form a compound with the general formula MgX. What would the most likely formula be f

ziro4ka [17]

Answer:K2X

Explanation: Valency can be defined as the combining power of an element. It is the valency that dictates the value an element will have when writing a chemical formula for its compound.

MgX is a compound of magnesium and an element X. The valency of magnesium in most of its compound is +2. Now for the 2 to have been absent in the chemical formula, this shows that the element X itself have a valency if -2 for the valencies of both to have canceled out.

Now considering the element potassium, it is an alkaline metal belonging to group 1 of the periodic table. Hence, it is expected that it has a valency of +1

Forming a compound with element X means there would be an exchange of valencies between the two. We have established that x has a valency of -2. The formula of the compound thus formed by exchanging the valencies of both element would be K2X

7 0

2 years ago

Read 2 more answers

A chemist measures the amount of hydrogen gas produced during an experiment. She finds that 264. g of hydrogen gas is produced.

Lesechka [4]

Answer:

The answer is 130.953 g of hydrogen gas.

Explanation:

Hydrogen gas is formed by two atoms of hydrogen (H), so its molecular formula is H₂. We can calculate is molecular weight as the product of the molar mass of H (1.008 g/mol):

Molecular weight H₂= molar mass of H x 2= 1.008 g/mol x 2= 2.01568 g

Finally, we obtain the number of mol of H₂ there is in the produced gas mass (264 g) by using the molecular weight as follows:

mass= 264 g x 1 mol H₂/2.01568 g= 130.9731703 g

The final mass rounded to 3 significant digits is 130.973 g

4 0

3 years ago

In an experiment, 107.9 grams of H2SO, is produced when 196.2 grams

Ipatiy [6.2K]

Answer:

54.99% yield

Explanation:

percent yield is just the amount you obtained over the amount expected times 100%.

(experimental value/theoretical value) x 100%

= (107.9 g/196.2 g) x 100%

=54.99% yield

5 0

3 years ago

what modifications to either or both of the half-cell reactions would kim need to make if she wanted to use them to produce equa

kompoz [17]

We have to add the both half cell equations and eliminate the number of electrons lost/gained.

<h3>What modification must Kim make to the equations?</h3>

The term redox reaction is a type of reaction that occurs when an electron is lost or gained in a reaction system. We can see that in this reaction, zinc looses two electron which are gained by copper.

If we want to obtain the equation 4.9 which is the overall equation of the redox reaction from the various half cell equations then we have to add the both half cell equations and eliminate the number of electrons lost/gained.

Learn kore about redox reaction:brainly.com/question/13293425

#SPJ1

7 0

2 years ago

Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly

trapecia [35]

Answer : The mass of silver sulfadiazine produced can be, 71.35 grams.

Solution : Given,

Mass of $Ag_2O$ = 25.0 g

Mass of $C_{10}H_{10}N_4SO_2$ = 50.0 g

Molar mass of $Ag_2O$ = 231.7 g/mole

Molar mass of $C_{10}H_{10}N_4SO_2$ = 250.3 g/mole

Molar mass of $AgC_{10}H_{9}N_4SO_2$ = 357.1 g/mole

First we have to calculate the moles of $Ag_2O$ and $C_{10}H_{10}N_4SO_2$ .

$\text{ Moles of }Ag_2O=\frac{\text{ Mass of }Ag_2O}{\text{ Molar mass of }Ag_2O}=\frac{25.0g}{231.7g/mole}=0.1079moles$

$\text{ Moles of }C_{10}H_{10}N_4SO_2=\frac{\text{ Mass of }C_{10}H_{10}N_4SO_2}{\text{ Molar mass of }C_{10}H_{10}N_4SO_2}=\frac{50.0g}{250.3g/mole}=0.1998moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Ag_2O(s)+2C_{10}H_{10}N_4SO_2(s)\rightarrow 2AgC_{10}H_9N_4SO_2(s)+H_2O(l)$

From the balanced reaction we conclude that

As, 2 mole of $C_{10}H_{10}N_4SO_2$ react with 1 mole of $Ag_2O$

So, 0.1998 moles of $C_{10}H_{10}N_4SO_2$ react with $\frac{0.1998}{2}=0.0999$ moles of $Ag_2O$

From this we conclude that, $Ag_2O$ is an excess reagent because the given moles are greater than the required moles and $C_{10}H_{10}N_4SO_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AgC_{10}H_9N_4SO_2$