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3 years ago

How many chlorine molecules are in 35.5g of chlorine gas

1 answer:

7 0

Since chlorine is one of the 7 diatomic elements we know that chlorine appears as Cl₂ gas naturally. That means that the molar mass of a chlorine gas is 70.9g/mol. That being said, first you need to find the number of moles of chlorine gas that are present in a 35.5g sample. To do this divide 35.5g by the molar mass of chlorine gas (70.9g/mol) to get 0.501mol of chlorine. Then you have to multiply 0.501mol by 6.02×10²³ to get the number of chlorine gas molecules. Therefore 3.01×10²³ molecules of chlorine gas are present in a 35.5g sample.

I hope that helps. Let me know in the comments if anything is unclear.

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A galvanic cell whose cell reaction is 2Fe3+(aq) + Zn(s) → 2Fe2+(aq) + Zn2+(aq) has a cell potential of 0.72V. What is the maxim

sesenic [268]

Answer:

138.96kJ is the maximum electrical work

Explanation:

The maximum electrical work that can be obtained from a cell is obtained from the equation:

W = -nFE

<em>Where W is work in Joules,</em>

<em>n are moles of electrons = 2mol e- because half-reaction of Zn is:</em>

Zn(s) → Zn²⁺(aq) + 2e⁻

F is faraday constant = 96500Coulombs/mol

E is cell potential = 0.72V

Replacing:

W = -2mol*96500Coulombs/mol*0.72V

W = - 138960J =

<h3>138.96kJ is the maximum electrical work</h3>

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2 years ago

What is the compound name for the ion chemical formula AgBr?

Gnom [1K]

Answer:

Silver bromide.

Explanation:

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3 years ago

If 12.1 kilograms of al2o3(s), 60.4 kilograms of naoh(l), and 60.4 kilograms of hf(g) react completely, how many kilograms of cr

Fynjy0 [20]

Answer is: 7,826 kg of cryolite.
Chemical reaction: Al₂O₃ + 6NaOH + 12HF → 2Na₃AlF₆ + 9H₂<span>O.
m(</span>Al₂O₃) = 12,1 kg = 12100 g.
n(Al₂O₃) = m(Al₂O₃) ÷ M(Al₂O₃).
n(Al₂O₃) = 12100 g ÷ 101,96 g/mol = 111,86 mol; limiting reactant.
m(NaOH) = 60,4 kg = 60400 g.
n(NaOH) = 60400 g ÷ 40 g/mol.
n(NaOH) = 1510 mol.
m(HF) = 60,4 kg = 60400 g.
n(HF) = 60400 g ÷ 20 g/mol = 3020 mol.
From chemical reaction: n(Al₂O₃) : n(Na₃AlF₆) = 6 : 2.
n(Na₃AlF₆) = 2 ·111,86 mol ÷ 6 = 37,28 mol.
m(Na₃AlF₆) = 37,28 mol · 209,94 g/mol.
m(Na₃AlF₆) = 7826,56 g = 7,826 kg.

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3 years ago

Question 14 A student dissolves 1.5g of styrene C8H8 in 225.mL of a solvent with a density of 1.02/gmL . The student notices tha

Yuki888 [10]

We must to know:

Cm = molarity = niu / Vs, when the niu = no. of moles and Vs = Volume of solution

the no. niu = mass / molecular mass of substance

molecular mass of C8H8 = 12x8+8x1 = 104 g/mol

=> niu = 1,5 / 104 = 0,0144 moles C8H8

=> Cm = 0,0144/0,225 = 0,06 mol/L

Cmm = molality = niu (C8H8) / mass of solvent (kg)

=> p = mass / V => mass (solvent) = p x V

=> 225 x 1,02 = 229,5 g solvent = 0,2295 kg solvent

=> Cmm = 0,0144 / 0,229,5 = 0,063

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3 years ago

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Kaylis [27]

the answer is d. when more solute is added

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3 years ago