1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Bess [88]
3 years ago
14

An 82.0 g sample of hydrated magnesium sulfate (MgSO4) is heated to constant mass of 40.1 grams. What is the formula for the hyd

rate?
Chemistry
1 answer:
Ket [755]3 years ago
8 0

Answer:

MgSO4.7H2O

Explanation:

The heating removes all the water (H2O) in the crystal.

The mass of the water removed = 82-40.1 = 41.9 grams.

The formula of the hydrated salt = MgSO4. xH2O

The ratio in of the masses of MgSO4 : H2O =  40.1 : 41.9

Dividing these by the molar masses we get

40.1/ 120.4 to  41.9 / 18.016

= 0.33 : 2.33

= 1 : 7.

So there are 7 water molecules n the hydrate.

You might be interested in
At a certain temperature and pressure, one liter of CO2 gas weighs 1.95 g.
AysviL [449]

Answer:

1.332 g.

Explanation:

  • We can use the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • At the same T and P and constant V (1.0 L), different gases have the same no. of moles (n):

<em>∴ (n) of CO₂ = (n) of C₂H₆</em>

<em></em>

∵ n = mass/molar mass

<em>∴ (mass/molar mass) of CO₂ = (mass/molar mass) of C₂H₆</em>

mass of CO₂ = 1.95 g, molar mass of CO₂ = 44.01 g/mol.

mass of C₂H₆ = ??? g, molar mass of C₂H₆ = 30.07 g/mol.

<em>∴ mass of C₂H₆ = [(mass/molar mass) of CO₂]*(molar mass) of C₂H₆</em> = [(1.95 g / 44.01 g/mol)] * (30.07 g/mol) =<em> 1.332 g.</em>

<em></em>

7 0
3 years ago
If you have 8.943 L of carbon monoxide gas at SATP, how many moles would it contain?
viktelen [127]
1mole contains 22.4Lmol^-1
xmole contains 8.943
cross-multiply
x=1×8.943/22.4
x=0.40mole
there it contains 0.40moles.
8 0
3 years ago
Hey hiihhbdndkdkdjdjddjdj
grigory [225]

Answer:

4 cups are about 0.25 gallons

Explanation:

To convert a cup measurement to a gallon measurement, divide the volume by the conversion ratio. The volume in gallons is equal to the cups divided by 16.

7 0
2 years ago
Read 2 more answers
I need help on this question
uranmaximum [27]

Answer:

When a front passes over an area, it means a change in the weather. Many fronts cause weather events such as rain, thunderstorms, gusty winds, and tornadoes. At a cold front, there may be dramatic thunderstorms. At a warm front, there may be low stratus clouds. Usually, the skies clear once the front has passed.

Explanation:

Plz make me brainliest :)

Have a nice day

6 0
3 years ago
Read 2 more answers
what is the volume of a sample of liquid mercury that has a mass of 76.2 g, given that the density of mercury is 13.6 g/ml?
lilavasa [31]
24 square root . i know im wrong but so what 
6 0
3 years ago
Other questions:
  • What mass of aluminum chloride could be made from 8.1 g of aluminum and 4.2 L of chlorine at STP?
    10·1 answer
  • I NEED HELPP<br> Describe how carbon and oxygen form carbon dioxide.
    12·2 answers
  • What is an example of a chemical change that happens inside your body?
    10·2 answers
  • If 4520 kj of heat is needed to boil a sample of water, what is the mass of water
    7·1 answer
  • I think that rock from the mantle is added to the edges of both plates at
    7·1 answer
  • NHa is an<br> example of<br> O an element.<br> O a mixture.<br> O a compound<br> O an atom.
    7·2 answers
  • How can scientists control the flammability of a substance?
    10·1 answer
  • Please refer to image please
    6·1 answer
  • What volume of 12M HCI is needed to prepare 250<br> of 0.20M HCI?
    13·1 answer
  • an atmosphere is considered hazardous if it contains a hazardous gas in excess of 10 percent of the hazardous material's:
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!