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3 years ago

11

an unknown compound of 9 moles of hydrogen, 3 moles of carbon, and 3 moles of bromine. The empirical formula of the compound sho

uld be?

1 answer:

e-lub [12.9K]3 years ago

3 0

Answer:

CH3Br ( Methyl Bromide )

Explanation:

H9 C3 Br3 => rewrite as: C3H9Br3 => actually, this is 3 moles of molecule CH3Br !

C3H9Br3 ~ 3( CH3Br)

I hope this answer can help you. Have a nice day!

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What tool do I need to use to find the molar mass?

Inga [223]

Answer:

I never thought of it as a tool, but I think you refer to Atomic Weight. The total atomic weight of the atoms in a molecule is its Mass in Moles.

Take a really simple example H2O. The atomic weight of Oxygen is 16, Hydrogen is 1, and so added, 1 + 1 + 16 = 18; the mass of 1 mole of water is 18g. Same for Benzene, C6H6; atomic weight of carbon is 12, so 6 x 12 = 72, add 6 x 1, or 6, and 72 + 6 = 78

Explanation: The sum of the atomic weights of the atoms in a molecule will give you the Molecular Weight, I.e. the Mass of 1 Mole of thst molecule.

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4 years ago

You placed 43.1 g of an unknown metal at 100 °C into a coffee cup calorimeter that contained 50.0 g of water that was initially

nika2105 [10]

Answer :

(a) The heat released by the metal is -312.48 J

(b) The specific heat of the metal is $0.0944J/g^oC$

Explanation :

<u>For part A :</u>

Heat released by the metal = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the metal

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $51.5J/^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_2$ = mass of water = 50.0 g

$\Delta T$ = change in temperature = $(T_{final}-T_{initial})=23.2-22.0=1.2^oC$

Now put all the given values in the above formula, we get:

$q=[(51.5J/^oC\times 1.2^oC)+(50.0g\times 4.184J/g^oC\times 1.2^oC)]$

$q=312.48J$

Thus, the heat released by the metal is -312.48 J

<u>For part B :</u>

$q=m\times c\times \Delta T$

q = heat released by the metal = -312.48 J

m = mass of metal = 43.1 g

c = specific heat of metal = ?

$\Delta T$ = change in temperature = $(T_{final}-T_{initial})=23.2-100=76.8^oC$

Now put all the given values in the above formula, we get:

$-312.48J=43.1g\times c\times 76.8^oC$

$c=0.0944J/g^oC$

Thus, the specific heat of the metal is $0.0944J/g^oC$

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Kipish [7]

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Setler79 [48]

Answer:

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