What reaction is 2HI --> H2 + 12

Which of the following is not present in an atom?<br> Electron<br> Element<br> Neutron<br> Proton

Licemer1 [7]

Answer:

an element is not in an atom

Explanation:

because the only things in a atom are electrons neutrons and protons

4 0

2 years ago

Read 2 more answers

a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re

Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Given : 59.4 g of $H_2SO_4$ in 100 g of solution

moles of $H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61$

Volume of solution = $\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.61\times 1000}{54.6ml}=11.2M$

To calculate the volume of acid, we use the equation given by neutralisation reaction:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of stock acid which is $H_2SO_4$

$M_2\text{ and }V_2$ are the molarity and volume of dilute acid which is $H_2SO_4$

We are given:

$M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL$

Putting values in above equation, we get:

$11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL$

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

4 0

2 years ago

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

2 years ago

Balance the equations

bearhunter [10]

1. 2 H2 + O2 = 2 H2O

2. 6 K + B2O3 = 3 K2O + 2 B

3. 10 Na + 2 NaNO3 = 6 Na2O + N2

6 0

2 years ago

What mass of H₂ is needed to react with 8.75 g of O₂ according to the following equation: O2(g) + H2(g) → H₂O(g)?

FromTheMoon [43]

Explanation:

For reacting with 8.75 grams of oxygen, 1.08 grams of hydrogen is required.

The given balanced equation has been:

\rm O_2\;+\;2\;H_2\;\rightarrow\;H_2OO2+2H2→H2O

From the equation, 1 mole of oxygen reacts with 2 mole of hydrogen to give 1 mole of water.

The mass of oxygen has been: 8.75 g,

Moles = \rm \dfrac{weight}{molecular\;weight}molecularweightweight

Moles of oxygen = \rm \dfrac{8.75}{32}328.75

Moles of oxygen = 0.27 mol

Since,

1 mole Oxygen = 2 mole hydrogen

0.21 mol oxygen = 0.54 mol hydrogen

Mass of hydrogen = moles \times× molecular weight

Mass of hydrogen = 0.54 \times× 2

Mass of hydrogen = 1.08 grams.

Thus, for reacting with 8.75 grams of oxygen, 1.08 grams of hydrogen is required.

6 0

1 year ago