If the cardboard square has a light on it what do observe

Manganese(iv) oxide reacts with aluminum to form elemental manganese and aluminum oxide: 3mno2+4al→3mn+2al2o3part awhat mass of

Oxana [17]

12.4 g First, calculate the molar masses by looking up the atomic weights of all involved elements. Atomic weight manganese = 54.938044 Atomic weight oxygen = 15.999 Atomic weight aluminium = 26.981539 Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol Now determine the number of moles of MnO2 we have 30.0 g / 86.936044 g/mol = 0.345081265 mol Looking at the balanced equation 3MnO2+4Alâ†’3Mn+2Al2O3 it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So 0.345081265 mol / 3 * 4 = 0.460108353 mol So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum. 0.460108353 mol * 26.981539 g/mol = 12.41443146 g Finally, round to 3 significant figures, giving 12.4 g

7 0

3 years ago

what is the molar mass of a gaseous flouride of sulfur containing 70.4% F and having a density of approximately 4.5g/L at 20 deg

zlopas [31]

Answer:

The molar mass is 180.2 g/mol

Explanation:

Step 1: Data given

% of F = 70.4 %

Density = 4.5 g/L

Temperature = 20 °C

Pressure = 1 atm

Step 2: Calculate the number of moles

PV = nRT

⇒ with P = the pressure = 1.00 atm

⇒ with V = the volume = Assume this is 1L

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

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

⇒ T = the temperature : 20°C = 293 Kelvin

1 atm*1L= n(0.08206 L-atm/mol-K)*(293 K)

n = 0.04159 moles

Step 3: Calculate molar mass

Molar mass = Mass / moles

4.5 grams / 0.04159 moles = 108.2 g/mol

Step 4: Calculate moles of F

Moles = Mass / molar mass

Moles F = 70.4 g / 19 g/mol

Moles F = 3.70 moles

Moles S = 29.6g / 32.07 g/mol

Moles S = 0.923 moles S

Step 5: Divide by the smallest amount of moles

F = 3.70 / 0.923 = 4

S = 0.923 / 0.923 = 1

The empirical formula is SF4

The molar mass of SF4 = 32.07 + 4*19 = 108.07 g/mol

This means the empirical formula is the same as the molecular formula SF4

The molar mass is 180.2 g/mol

4 0

3 years ago

Draw the Lewis dot structure for RbIO2. Include all hydrogen atoms and nonbonding electrons. Show the formal charges of all atom

Rzqust [24]

Answer : The Lewis-dot structure of $RbIO_2$ is shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $RbIO_2$

As we know that rubidium has '1' valence electrons, iodine has '7' valence electrons and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in $RbIO_2$ = 1 + 7 + 2(6) = 20

As we know that $RbIO_2$ is an ionic compound because it is formed by the transfer of electron takes place from metal to non-metal element.

7 0

3 years ago

Below is a word equation. Please follow directions for each part below.

Pavel [41]

HCl + NaOH ---> NaCl + H20

If you follow guidance from other questions I have already answered for you, you will see that the above equation is balanced as it is.

4 0

2 years ago

according to Boyles law if the initial pressure and volume of a gas were five atm for the pressure and 10 mL for the volume and

maks197457 [2]

Answer:The new volume is 5mL

Explanation:

The formular for Boyles Law is; P1 V1 = P2 V2

Where P1 = 1st Pressure V1 = First Volume

P2 = 2nd Pressure V2 = Second Volume

From the question; P1 = 5atm, V1 = 10ml

P2 = 2 x P1 (2 x 5) = 10 atm V2 =?

Using the Boyles Law Formular; P1 V1 = P2 V2, we make V2 the subject of formular; P1 V1/ P2 = V2

∴ 5 x 10/ 10 = 5

∴ V2 = 5mL

5 0

3 years ago

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If the cardboard square has a light on it what do observe​

If the cardboard square has a light on it what do observe