Answer : The density of a sample of chlorine gas is, 12.59 g/L
Explanation :
To calculate the density of chlorine gas we are using ideal gas equation.

where,
P = pressure of chlorine gas = 4.5 atm
V = volume of chlorine gas = 12.6 L
n = number of moles of chlorine gas
w = mass of chlorine gas
R = gas constant = 0.0821 L.atm/mol.K
T = temperature of chlorine gas = 
M = molar mass of chlorine gas = 71 g/mol
Now put all the given values in the above formula, we get:



Therefore, the density of a sample of chlorine gas is, 12.59 g/L
Answer: -
3.3° C
Explanation: -
Mass of water m = 180.5 g
Energy released as heat Q = 2494 J
Specific heat is defined as the heat required to raise the temperature of the unit mass of a given substance by 1 C.
Specific heat of water Cp = 4.184 (J/g)⋅∘C
Using the formula
Q = m x Cp x ΔT
We get temperature change ΔT = Q / (m x Cp)
= 2494 J / ( 180.5 g x 4.184 (J/g)⋅∘C
= 3.3° C
Thus the temprature change, (ΔT), of the wateris 3.3 °C if 180.5 g of water sat in the copper pipe from part A, releasing 2494 J of energy to the pipe
Answer:
B)3
Explanation:
I'm assuming the 2 is meant to be smaller. If it were a regular size 2 the answer would be 2 and that isn't an option
<span>convert grams to moles then use the equatino to do moles of Na to moles of H2O then convert moles of H2O to molecules by using avogadros number (6.022e23)
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Answer:
In the given chemical reaction:
Species Oxidized: I⁻
Species Reduced: Fe³⁺
Oxidizing agent: Fe³⁺
Reducing agent: I⁻
As the reaction proceeds, electrons are transferred from I⁻ to Fe³⁺
Explanation:
Redox reaction is a chemical reaction involving the simultaneous movement of electrons thereby causing oxidation of one species and reduction of the other species.
The chemical species that <u><em>gets reduced by gaining electrons </em></u><u>is called an </u><u><em>oxidizing agent</em></u>. Whereas, the chemical species that <u><em>gets oxidized by losing electrons </em></u><u>is called a </u><u><em>reducing agent</em></u><u>.</u>
Given redox reaction: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
<u>Oxidation half-reaction</u>: 2 I⁻ + → I₂ + 2 e⁻ ....(1)
<u>Reduction half-reaction</u>: [ Fe³⁺ + 1 e⁻ → Fe²⁺ ] × 2
⇒ 2 Fe³⁺ + 2 e⁻ → 2 Fe²⁺ ....(2)
In the given redox reaction, <u>Fe³⁺ (oxidation state +3) accepts electrons and gets reduced to Fe²⁺ (oxidation state +2) and I⁻ (oxidation state -1) loses electrons and gets oxidized to I₂ (oxidation state 0).</u>
<u>Therefore, Fe³⁺ is the oxidizing agent and I⁻ is the reducing agent and the electrons are transferred from I⁻ to Fe³⁺.</u>