M = Sr (strontium)
<em>Step 1.</em> Calculate the <em>moles of CO_2</em>.
Moles of CO_2 = 0.395 g CO_2 × (1 mol CO_2/44.01 g CO_2)
= 0.008 975 mol CO_2
<em>Step 2</em>. Calculate the <em>moles of MCO_3</em>.
Moles of MCO_3 = 0.008 975 mol CO_2 × (1 mol MCO_3/1 mol CO_2)
= 0.008 975 mol MCO_3
<em>Step 3</em>. Calculate the molar mass of <em>MCO_3</em>
MM = grams/moles = 1.324 g/0.008 75 mol = 147.5 g/mol
<em>Step 4</em>. Calculate the <em>atomic mass of M</em>
M_r = <em>x</em> + 12.01 + 3×16.00 = <em>x</em> + 60.01 = 147.5
<em>x</em> = 147.5 – 60.01 = 87.5
<em>Step 5. Identify M</em>.
The element with the closest atomic mass is Sr (A_r = 87.6).
∴ M = Sr and the compound is SrCO_3.
Answer:
The second experiment (reversible path) does more work
Explanation:
Step 1:
A piston confines 0.200 mol Ne(g) in 1.20L at 25 degree °C
<em>(a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm</em>
<em></em>
Irreversible path: w =-Pex*ΔV
⇒ with Pex = 1.00 atm
⇒ with ΔV = 1.20 L
W = -(1.00 atm) * 1.20 L
W = -1.20L*atm *101.325 J /1 L*atm = -121.59 J
<em>(b) The gas is allowed to expand reversibly and isothermally to the same final volume.</em>
<em></em>
W = -nRTln(Vfinal/Vinitial)
⇒ with n = the number of moles = 0.200
⇒ with R = gas constant = 8.3145 J/K*mol
⇒ with T = 298 Kelvin
⇒ with Vfinal/Vinitial = 2.40/1.20 = 2
W = -(0.200mol) * 8.3145 J/K*mol *298K *ln(2.4/1.2)
W = -343.5 J
The second experiment (reversible path) does more work
As,
5471 kJ heat is given by = 1 mole of Octane
Then,
5310 kJ heat will be given by = X moles of Octane
Solving for X,
X = (5310 kJ × 1 mol) ÷ 5471 kJ
X = 0.970 moles of Ocatne
So, 0.970 moles of Octane will liberate 5310 kJ energy. Now changing moles to mass,
As,
Moles = mass / M.mass
Or,
Mass = Moles × M.mass
Putting values,
Mass = 0.970 mol × 114.23 g/mol
Mass = 110.83 g of Octane
The answer is: The engine will run inefficiently APEX....
Answer: <span>A-Ce is oxidized because it is losing electrons and Cu is reduced because it is gaining electrons</span><span>.
</span>There are two reactions in the equation, oxidation and reduction. A molecule that oxidized will lose electrons while the molecule that reduced will gain electrons. In this case, Cu2+ changed into Cu which means its oxidation number reduced from +2 into 0. Ce oxidation number increased from 0 into +3
