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frez [133]
3 years ago
11

What type of extinguisher is filled with a fine powder

Chemistry
2 answers:
PolarNik [594]3 years ago
5 0
The ABC fire extinguisher.
Marysya12 [62]3 years ago
4 0
I believe it is the abc fire exstignusher
You might be interested in
What is reduction potential?​
Shtirlitz [24]

Answer:

A reduction potential measures the tendency of a molecule to be reduced by taking up new electrons. ... Standard reduction potentials can be useful in determining the directionality of a reaction. The reduction potential of a given species can be considered to be the negative of the oxidation potential.

Explanation:

8 0
2 years ago
Read 2 more answers
Answers to all of these
Paul [167]

Answer:

1. Percent composition of  Al = 13.423 %

2.

  • Percent composition of Zn = 28.02 %
  • Percent composition of Cl = 30.6 %
  • Percent composition of O = 41.3 %

3. The empirical formula is C₅O₁₆

4. Molecular Formula= P₄O₆

Explanation:

Part first :

Data Given

Formula of the Molecule = Al₂ (CrO₄)₃

% of Al₂ = ?

> First of all find the atomic masses of each component in a molecule

For Al₂ (CrO₄)₃ atomic masses are given below

Al = 27 g/mol

Cr = 52 g/mol

O = 16 g/mol

> Then find the total masses of each component

2 atoms of Al = 27 g/mol x 2

= 54 g/mol

3 atoms of Cr = 52 g/mol x 3

= 156 g/mol

12 atoms of O = 16 g/mol x 12

= 192 g/mol

> find total Molar Mass of Molecule:

Molar Mass of Al₂ (CrO₄)₃ = [27x2 + 52x3 + 16x12]

Molar Mass of Al₂ (CrO₄)₃ = 402

Now to find the mass percent of Al

Formula used to find the Mass percent of a component

Percent composition of  Al = mass of Al in Molecula / molar mass of Al₂(CrO₄)₃ x 100%

Put the values

Percent composition of  Al =  54 (g/mol) / 402 (g/mol) x 100%

Percent composition of  Al = 13.423 %

_______________________________________

Part 2

Data Given

Formula of the Molecule = Zn(ClO₃)₂

% Zn = ?

% Cl = ?

% O = ?

> First of all find the atomic masses of each component in a molecule

For Zn(ClO₃)₂ atomic masses are given below

Zn = 65 g/mol

Cl = 35.5 g/mol

O = 16 g/mol

> Then find the total masses of each component

1 atoms of Zn= 65 g/mol x 1

= 65 g/mol

2 atoms of Cl = 35.5 g/mol x  

= 71 g/mol

6 atoms of O = 16 g/mol x 6

= 96 g/mol

> find total Molar Mass of Molecule:

Molar Mass of Zn(ClO₃)₂ = [65x1 + 35.5x2 + 16x6]

Molar Mass of Zn(ClO₃)₂ = 232g/mol

Now to find the mass percent of of each component one by one

1.  Formula used to find the mass percent of Zn

Percent composition of  Zn= mass of Zn in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of Zn = 65(g/mol) / 232 (g/mol) x 100%

Percent composition of Zn = 28.02 %

-------------------

2.  Formula used to find the mass percent of Cl

Percent composition of  Cl = mass of Cl in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of Cl = 71 (g/mol) / 232 (g/mol) x 100%

Percent composition of Cl = 30.6 %

---------------------

3.  Formula used to find the mass percent of O

Percent composition of  O = mass of O in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of O = 96 (g/mol) / 232 (g/mol) x 100%

Percent composition of O = 41.3 %

________________________________________

Part 3:

Data Given

Percentage of C = 27.3 %

Percentage of O = 72.7 %

Emperical Formula of the compound = ?

Solution:

So the compound has 27.3 % C and 72% O

First, find the mass of each of the elements in 100 g of the Compound.

C = 27.3 g

O = 72 g

Now find how many moles are there for each element in 100 g of compound

For this molar mass are required

That is

C = 12 g/mol

O = 16 g/mol

Formula Used

mole of C = mass of C / Molar mass of C

 mole of C = 27.3 / 12 g/mol

  mole of C = 2.275

Formula Used

mole of O = mass of O / Molar mass of O

 mole of O = 72g / 16 g/mol

  mole of O = 7.2

Divide each one by the smallest number of moles

C = 2.275 / 2.275

C = 1

O = 7.2 / 2.275

O = 3.2

Multiply the mole fraction to a number to get the whole number.

C = 1 x 5 = 5

O = 3.2 x 5 =  16

So, the empirical formula is C₅O₁₆

______________________________________

Part 4

Data Given

Percentage of P= 56.38 %

Percentage of O = 43.62%

Molar Mass = 219.9g

Molecular Formula of the compound = ?

Solution:

First, find the mass of each of the elements in 100 g of the Compound.

Mass of P= 56.38g

Mass of O = 43.62g

Now find how many moles are there for each element in 100 g of compound

find the moles in total compounds

Formula Used

mole of P = mass of  / Molar mass of P

 mole of P = 56.38 g / 31 g/mol

  mole of P = 1.818

Formula Used

mole of O = mass of O / Molar mass of O

 mole of O = 43. 62 / 16 g/mol

  mole of O = 2.7262

Now

first find the Emperical formula

Divide each one by the smallest number of moles

P = 1.818 /1.818

P= 1

for oxygen

O = 2.7262 / 1.818

O = 1.5

Multiply the mole fraction to a number to get the whole number.

P = 1 x 2 = 2

O = 1.5 x 2 =  3

So, the empirical formula is P₂O₃

Now  

Find molar mass of the empirical formula P₂O₃

2 (31) + 3 (16) = 62 + 48 = 110

Now find that how many empirical units are in a molecular unit.

(219.9 g/mol) / ( 110 g/mol) =  empirical units per molecular unit

empirical units per molecular unit = 1.999 =2

A here we get two empirical units in a molecular unit,

So the molecular formula is:

2 (P₂O₃) = P₄O₆

7 0
3 years ago
How many is the maximum number of electrons for n=4
mihalych1998 [28]
In Chemistry, to better determine the position of a certain electron, quantum numbers are used. The four quantum numbers are n, l, m and s. In the given above of n= 4, the principal quantum number is 4 and this represents the overall relative energy in the orbital. This means that we are to find the maximum number of electrons in fourth main energy level and the answer is 32. 
7 0
3 years ago
Under which conditions will sugar most likely dissolve fastest in a cup of water
qwelly [4]
It will dissolve fastest in a cup of hot water. 
4 0
3 years ago
Read 2 more answers
How many kilojoules of heat will be released when exactly 1 mole of iron, Fe, is burned to form Fe2O3 at standard state condaiti
Diano4ka-milaya [45]

Answer:

412.1kJ

Explanation:

For the reaction , from the question -

4Fe (s)  +  3O₂ (g)  → 2Fe₂O₃ (s)

Δ Hrxn = Δ H°f (products) - Δ H°f (reactants)

In case the compound is in its standard state , enthalphy of formation is zero

Hence ,

for the above reaction ,

ΔHrxn =( 2 * Δ H° (Fe₂O₃ )) - [ ( 4 *Δ H° Fe ) + (3 * Δ H° O₂ )]

The value for Δ H°(Fe₂O₃ ) = - 824.2kJ/mol

Δ H° Fe = 0

Δ H° O₂ = 0

Putting in the above equation ,

ΔH rxn = ( 2 * Δ H° (Fe₂O₃ ))  - 0

ΔHrxn =  2× - 824.2 kJ / mol = - 1648.4 kJ/mol

- 1648.4 kJ/mol  , this much heat is released by the buring of 4 mol of Fe.

Hence ,

for 1 mol of Fe ,

- 1648.4 kJ/mol  / 4 = 412.1kJ

8 0
3 years ago
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