Which of the subatomic particles is responsible for the chemical behavior of a given atom?

How many milligrams of magnesium reacts with excess HCl to produce 31.2 mL of hydrogen gas at 754 Torr and 25.0℃.

Ivanshal [37]

Answer:

There will react 30.9 milligrams of magnesium

Explanation:

Step 1: Data given

Volume of hydrogen = 31.2 mL

Pressure = 754 torr = 754/760 = 0.992 atm

Temperature = 25.0 °C = 298 Kelvin

Step 2: The balanced equation

Mg + 2HCl → MgCl2 + H2

Step 3: Calculate moles of H2

p*V = n*R*T

⇒ with p = the pressure of H2 = 0.992 atm

⇒with V = the volume of H2 = 31.2 mL = 0.0312 L

⇒ with n = the moles of H2 = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T= the temperature = 25.0 °C = 298 Kelvin

n = (p*V)/(R*T)

n = (0.992*0.0312)/(0.08206*298)

n = 0.00127 moles

Step 4: Calculate moles of Mg

For 1 mol of Mg we need 2 moles of HCl to produce 1 mol of MgCl2 and 1 mol of H2

For 0.00127 moles of H2 we need 0.00127 moles of Mg

Step 5: Calculate mass of Mg

Mass of Mg = moles of Mg * molar mass of Mg

Mass of Mg = 0.00127 moles * 24.3 g/mol

Mass of Mg = 0.0309 grams = 30.9 mg of Mg

There will react 30.9 milligrams of magnesium

3 0

3 years ago

Which of the following is a scientific question? O A. How can I make a peach cobbler? O B. What chemicals cause most plants to b

igor_vitrenko [27]

The answer is B. Because think about it, it’s not a scientific fact that brown dogs are better pets, and the best color for your room isn’t it because the answer would be an opinion of someone not a fact. I hope this might help you.

5 0

3 years ago

The iodine "clock reaction" involves the following sequence of reactions occurring in a reaction mixture in a single beaker. 1.

Mars2501 [29]

C: 0.012 mol.

<h3>Explanation</h3>

Start with 0.0020 moles of iodate ions ${\text{IO}_{3}}^{-}$ .

How many moles of iodine $\text{I}_2$ will be produced?

${\text{IO}_{3}}^{-}$ converts to $\text{I}_2$ in the first reaction. The coefficient in front of $\text{I}_2$ is three times the coefficient in front of ${\text{IO}_{3}}^{-}$ . In other words, each mole of ${\text{IO}_{3}}^{-}$ will produce three moles of $\text{I}_2$ . 0.0020 moles of ${\text{IO}_{3}}^{-}$ will convert to 0.0060 moles of $\text{I}_2$ .

How many moles of thiosulfate ions ${\text{S}_2\text{O}_3}^{2-}$ are required?

$\text{I}_2$ reacts with ${\text{S}_2\text{O}_3}^{2-}$ in the second reaction. The coefficient in front of $\text{I}_2$ is twice the coefficient in front of ${\text{S}_2\text{O}_3}^{2-}$ . How many moles of ${\text{S}_2\text{O}_3}^{2-}$ does each mole of $\text{I}_2$ consume? Two. 0.0060 moles of $\text{I}_2$ will be produced. As a result, $2 \times 0.0060 = 0.0120$ moles of ${\text{S}_2\text{O}_3}^{2-}$ will be needed.

6 0

3 years ago

How many grams of NaOH needed to completely neutralize 3L of 1.75M HCL

Sedaia [141]

Answer:

209.98 g of NaOH

Explanation:

We are given;

Volume of HCl as 3 L
Molarity of HCl as 1.75 M

We are required to calculate the mass of NaOH required to completely neutralize the acid given.

First, we write a balanced equation for the reaction between NaOH and HCl

That is;

NaOH + HCl → NaCl + H₂O

Second, we determine the number of moles of HCl

Number of moles = Molarity × Volume

= 1.75 M × 3 L

= 5.25 moles

Third, we use the mole ratio to determine the moles of NaOH

From the reaction,

1 mole of NaOH reacts with 1 mole of HCl

Therefore;

Moles of NaOH = Moles of HCl

= 5.25 moles

Fourth, we determine the mass of NaOH

Molar mass of NaOH = 39.997 g/mol

Mass of NaOH = 5.25 moles × 39.997 g/mol

= 209.98 g

Thus, 209.98 g of NaOH will completely neutralize 3L of 1.74 M HCl

6 0

3 years ago

nguyên tử sắt có điện tích hạt nhân là 26+. trong nguyên tử, số hạt mang điện nhiều hơn số hạt không mang điện là 22. Hãy xác đị

dimaraw [331]

Answer:

I don't know yr language

8 0

3 years ago