Street: 1. Cars 2. People 3. Wind in the trees 4. Birds 5. Car horn
School: 1. School bell 2. Students laughing 3. Teacher talking 4. Footsteps 5. Lockers
Church: 1. Priest talking 2. People talking 3. Church music 4. Kids crying 5. Benches creaking
Answer:
22.77 g.
he limiting reactant is O₂, and the excess reactant is Mg.
Explanation:
- From the balanced reaction:
<em>Mg + 1/2O₂ → MgO,</em>
1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.
- We need to calculate the no. of moles of (16.3 g) of Mg and (4.52 g) of oxygen:
no. of moles of Mg = mass/molar mass = (16.3 g)/(24.3 g/mol) = 0.6708 mol.
no. of moles of O₂ = mass/molar mass = (4.52 g)/(16.0 g/mol) = 0.2825 mol.
So. 0.565 mol of Mg reacts completely with (0.2825 mol) of O₂.
<em>∴ The limiting reactant is O₂, and the excess reactant is Mg (0.6708 - 0.565 = 0.1058 mol).</em>
<u><em>Using cross multiplication:</em></u>
1.0 mole of Mg produce → 1.0 mol of MgO.
∴ 0.565 mol of Mg produce → <em>0.565 mol of MgO.</em>
<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.565 mol)(40.3 g/mol) = <em>22.77 g.</em>
Answer:
The maximum amount of work that can be done by this system is -2.71 kJ/mol
Explanation:
Maximum amount of work denoted change in gibbs free energy 
during the reaction.
Equilibrium concentration of B = 0.357 M
So equilibrium concentration of A = (1-0.357) M = 0.643 M
So equilibrium constant at 253 K, ![K_{eq}= \frac{[B]}{[A]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%20%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D)
[A] and [B] represent equilibrium concentrations
When concentration of A = 0.867 M then B = (1-0.867) M = 0.133 M
So reaction quotient at this situation, 
We know, 
where R is gas constant and T is temperature in kelvin
Here R is 8.314 J/(mol.K), T is 253 K, Q is 0.153 and 
is 0.555
So, 
= -2710 J/mol
= -2.71 kJ/mol
M=f/a m=129000N/84m/s/s=1535.714286kg/m/s/s
To solve this we assume
that the gas inside the balloon is an ideal gas. Then, we can use the ideal gas
equation which is expressed as PV = nRT. At a constant pressure and number of
moles of the gas the ratio T/V is equal to some constant. At another set of
condition of temperature, the constant is still the same. Calculations are as
follows:
T1 / V1 = T2 / V2
V2 = T2 x V1 / T1
V2 = 293.15 x 300 / 333.15
V2 = 263.98 cm^3
