All categories

4 years ago

Calculate The molar mass of aluminum nitrate Al(NO3)3

Chemistry

1 answer:

sleet_krkn [62]4 years ago

7 0

Answer:

Molar mass: 212.996 g/mol

Explanation:

Formula: Al(NO3)3

Molar mass: 212.996 g/mol

IUPAC ID: Aluminium nitrate

Melting point: 72.8 °C

Boiling point: 135 °C

Density: 1.72 g/cm³

Soluble in: Water, Alcohol

You might be interested in

=> The approximate mss of an electron is 1/10²⁷ g , Calculate the uncertainty in its velocity if the uncertainty in its posit

nasty-shy [4]

Answer:

Δx(m.Δv)=h/4π

here ,

Δv = uncertainty in velocity

10−11×10−27×Δv=6.626×10−34/4×22/7

=5.25×103ms−1

3 0

1 year ago

Read 2 more answers

Help!!!!!!!!!!!!!!!!!!

skad [1K]

The density of the sample : 0.827 g/L

<h3>Further explanation</h3>

In general, the gas equation can be written

$\large {\boxed {\bold {PV = nRT}}}$

where

P = pressure, atm , N/m²

V = volume, liter

n = number of moles

R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m2, v= m³)

T = temperature, Kelvin

n= 1 mol

MW Neon = 20,1797 g/mol

mass of Neon :

$\tt mass=mol\times MW\\\\mass=1\times 20,1797 =20.1797~g$

The density of the sample :

$\tt \rho=\dfrac{m}{V}\\\\\rho=\dfrac{20,1797}{24.4}=0.827~g/L$

or We can use the ideal gas formula ta find density :

$\tt \rho=\dfrac{P\times MW}{RT}\\\\\rho=\dfrac{1\times 20.1797}{0.082\times 298}\\\\\rho=0.826~g/L$

8 0

3 years ago

What temperature would be required to keep 0.895mol of gas in a 28.6L tank at a pressure of 0.500atm?

True [87]

Answer:

The temperature would be 194, 8 K.

Explanation:

We use the formula:

PV=nRT --> T=PV/nR

T= 0,500 atm x 28, 6 L/ 0,895 mol x 0,082 l atm /K mol

3 0

4 years ago

Question 1 (1 point)

lilavasa [31]

Answer:

I believe the answer is C. Stopping an object from moving

4 0

3 years ago

A flame test of a colorless solution gives a bright yellow color. When reacted with AgNO3 a white precipitate forms that dissolv

hichkok12 [17]

Answer:

The compound is Na2CO3

Explanation:

A flame test of a colorless solution gives a bright yellow color. The yellow color shows the presence of Sodium.

NaX + AgNO3 → AgX + NaNO3

When reacted with AgNO3 a white precipitate.

AgX + HNO3 → AgNO3 + H2O + CO2

A white precipitate will be formed when AgNO3 reacts with Sodium Chloride (NaCl) or Sodium carbonate(Na2CO3). (Also NaF, Na3PO4 are possible).

When HCL is added to the unknown solution, bubbles form.

Forming bubbles means that there is formed CO2 (and H20)

Na2X + 2HCl → 2NaCl + H2O + CO2

To form CO2 there is carbon needed

This shows that X is CO3 and the compound is Na2CO3

<u>To control:</u>

Na2CO3 + AgNO3 → Ag2CO3 + NaNO3

⇒Ag2Co3 is the white precipitate formed

This precipitate Ag2CO3 + HNO3 will disolve in AgNO3, CO2 and H2O

When Na2CO3 reacts with HCl there is formed NaCL together with bubbles ( which is CO2 and H2O).

The compound is Na2CO3

3 0

3 years ago