D) 710 g
Step by Step:
Multiply 2.5 L by 2.0 M to solve for moles
2 mol/L • 2.5 L= 5 mol
Find formula weight of sodium sulfate
Na2SO4-142.04 g/mol
Na- 2(22.99)=45.98
S-32.06
O-4(16)=64
Multiply miles by formula weight
5 mole • 142.04 g/mol=710.2 g
The answer is 267.93 g
Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol
The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%
Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
Answer:
the mass of water heated is m water = 18.81 g
Explanation:
Assuming a specific heat of iron of C iron =0.450 J/g°C (from tables) , then if we neglect heat losses to the environment , all the heat absorbed by the water is the one released by the iron ball.
Therefore
heat released= Q = m iron * C iron * (T initial iron - T final ) = 5 g * 0.450 J/g°C ( 200°C - 25°C) = 393.75 J
where
m= mass , T= temperature
then
heat released= heat absorbed =Q = m water * C water * (T final - T initial water)
m water = Q/ [C water * (T final - T initial water)] = 393.75 J/[ 4.186 J/g°C *( 25°C-20°C)] = 18.81 g
m water = 18.81 g
Moles of electrons:
The moles of electrons that are transferred are 12F
A balanced equation:
2 moles of Aluminium metal react with excess copper(II) nitrate.
Given:
Moles of Aluminium = 2
As Aluminium goes from 0 to +3 oxidation state
And copper goes from +2 to 0
On balancing the number of electrons we get:
For 1 mole of Al is required.
Therefore for 2 moles of Al,
Total F mole of electrons
Where F= Faraday's constant= 96500 C
So, 12F moles of electrons are transferred.
Learn more about Faraday's Law here,
brainly.com/question/27985929
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Assuming the molar mass of oxygen is 18.01, you only need 16 grams (B) of oxygen because the molar mass of hydrogen is about 1 gram.
16+1+1=18